(a) To estimate the value of f(2.5) using linear approximation, we can use the formula: f(x) ≈ f(x₀) + f'(x₀)(x - x₀). Given x₀ = 2, f(2) = 1, and f'(2) = 3, we can substitute these values into the formula:
f(2.5) ≈ f(2) + f'(2)(2.5 - 2).
f(2.5) ≈ 1 + 3(0.5).
f(2.5) ≈ 1 + 1.5.
f(2.5) ≈ 2.5.
Therefore, using linear approximation, we estimate that f(2.5) is approximately 2.5.
(b) To find a new estimate to a root of f(x) using one iteration of Newton's Method, we use the formula:
x₁ = x₀ - f(x₀)/f'(x₀).
Given x₀ = 2, we substitute this into the formula along with f(x₀) = 1 and f'(x₀) = 3:
x₁ = 2 - 1/3.
x₁ = 2 - 1/3.
x₁ = 5/3.
Therefore, one iteration of Newton's Method yields a new estimate to a root of f(x) as x₁ = 5/3.
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Which distribution has the largest median?
Group of answer choices
Set A
Set B
Set C
The distribution in Set C has the largest median.
The median of a distribution represents the middle value when the data points are arranged in ascending or descending order. To determine which distribution has the largest median, we need to compare the medians of Sets A, B, and C.
Without specific values or additional information about the sets, we cannot perform precise calculations or make a quantitative comparison. However, based on the available information, we can still provide a general answer.
Since the question asks about the distribution with the largest median, we can reason that the distribution in Set C has the largest median. This is because the question does not provide any indication or criteria that suggest otherwise.
Based on the given information and the question, we can conclude that the distribution in Set C has the largest median.
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question b and c
b. How many even numbers are between 1 and 101 , inclusive? c. How many multiples of 3 are between 1 and 101 , inclusive?
b. There are 51 even numbers between 1 and 101, inclusive.
c. There are 34 multiples of 3 between 1 and 101, inclusive.
b. An even number is divisible by 2. To find the number of even numbers between 1 and 101 (inclusive), we can divide the range by 2. The first even number in this range is 2, and the last even number is 100.
We can observe that there is a one-to-one correspondence between the even numbers and the counting numbers from 1 to 51.
Therefore, the number of even numbers in the given range is equal to the number of counting numbers from 1 to 51, which is 51.
c. A multiple of 3 is a number that can be evenly divided by 3. To find the number of multiples of 3 between 1 and 101 (inclusive), we divide the range by 3.
The first multiple of 3 in this range is 3, and the last multiple of 3 is 99. We can observe that there is a one-to-one correspondence between the multiples of 3 and the counting numbers from 1 to 34.
Therefore, the number of multiples of 3 in the given range is equal to the number of counting numbers from 1 to 34, which is 34.
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All the provided information is there please let me know what information you need.
Can you please answer 6,7,8,9,10
thank you
Molybdenum resistivity p= 5.34. (temp) a= .004579 1/C
radius r0 for your wire =3.58 mm
R0=11.59 mOhms
(6) then calculate how long l0 the wire must be, given its resistance R0. Tip: convert r0 to meters before you do anything else. Checks: the wire should be between 10 cm and 50 m long. Does l0 give the right resistance?
(7) Version: you also have a second wire, identical in every way to the original wire, except that its radius r is 45.3 % larger than the original wire. Example: to make the radius r larger than r0 by 31.1%, use r = 1.311 r0. To make r just 31.1% smaller than r0, it'd be: r = (1 – 0.311) r0 = 0.689 r0.
8. ) Calculate the second wire's resistance R in mΩ. Check: R can't be more than 4x larger/smaller than R0.
9. ) Now suppose you wanted to heat up or cool down the original wire so that R0 became equal to R, the resistance of the second wire. Would the original wire have to be heated or cooled? Explain without eqns.
10. Assuming that the both wires are initially at T0 = 20 °C, calculate the final temperature T of the original wire when its resistance is equal to R, the resistance of the second wire. Checks: does T bear out your prediction in Q10? Does T give the right R? Lastly, a tiny temperature change can't cause a large change in resistance.
The change in resistance is directly proportional to the change in temperature, a small temperature change should result in a small change in resistance.
Here are the answers to questions 6, 7, 8, 9, and 10 :
(6) To calculate the length [tex]$l_0$[/tex] of the wire given its resistance [tex]$R_0$[/tex], we can use the formula:
[tex]\[ R = \frac{{p \cdot l}}{{A}} \][/tex]
where R is the resistance, p is the resistivity, l is the length of the wire, and A is the cross-sectional area of the wire.
Given:
- Resistivity, p = 5.34
- Radius, [tex]$r_0 = 3.58 \, \text{mm} = 0.00358 \, \text{m}$[/tex]
- Resistance, [tex]$R_0 = 11.59 \, \text{m}\Omega = 0.01159 \, \Omega$[/tex]
First, we need to calculate the cross-sectional area of the wire:
[tex]\[ A = \pi \cdot r_0^2 \][/tex]
Substituting the values:
[tex]\[ A = \pi \cdot (0.00358)^2 \][/tex]
Next, we rearrange the resistance formula to solve for the length l:
[tex]\[ l = \frac{{R \cdot A}}{{p}} \][/tex]
Substituting the given values:
[tex]$\[ l_0 = \frac{{0.01159 \cdot \pi \cdot (0.00358)^2}}{{5.34}} \][/tex]
Evaluating the expression, we find:
[tex]\[ l_0 \approx 0.0000806 \, \text{m} \][/tex]
So, the length [tex]$l_0$[/tex] of the wire must be approximately 0.0000806 meters.
The wire length falls within the specified range of [tex]$10 \, \text{cm}$[/tex] and [tex]$50 \, \text{m}$[/tex], and the calculated resistance [tex]$R_0$[/tex] matches the given value.
(7). For the second wire, the radius r is 45.3% larger than the original wire's radius [tex]$r_0$[/tex].
We can calculate the new radius r using the formula:
[tex]\[ r = (1 + 0.453) \cdot r_0 \][/tex]
Substituting the given value:
[tex]\[ r = (1 + 0.453) \cdot 0.00358 \][/tex]
Calculating the expression:
[tex]\[ r \approx 0.00521 \, \text{m} \][/tex]
So, the radius of the second wire is approximately 0.00521 meters.
(8). To calculate the resistance R of the second wire, we use the same resistance formula:
[tex]\[ R = \frac{{p \cdot l}}{{A}} \][/tex]
We already know the resistivity p and the length l from the previous calculations.
We need to find the cross-sectional area A for the new radius r:
[tex]\[ A = \pi \cdot r^2 \][/tex]
Substituting the given values:
[tex]\[ A = \pi \cdot (0.00521)^2 \][/tex]
Calculating the expression:
[tex]\[ A \approx 0.00852 \, \text{m}^2 \][/tex]
Now, we can calculate the resistance R:
[tex]$\[ R = \frac{{5.34 \cdot 0.0000806}}{{0.00852}} \][/tex]
Calculating the expression:
[tex]R \approx 0.0506\ \Omega[/tex]
So, the resistance of the second wire, R, is approximately 0.0506, [tex]\Omega$ or $50.6 \, \text {m}\Omega$.[/tex]
The calculated resistance falls within the given check that R can't be more than 4 times larger or smaller than [tex]$R_0$[/tex].
(9). If we want to heat up or cool down the original wire (wire with resistance [tex]$R_0$[/tex]) to make its resistance equal to the resistance of the second wire (R), the original wire would need to be heated.
Heating the wire would increase its temperature, which in turn increases its resistance. By increasing the temperature, we can adjust the resistance of the original wire to match the resistance of the second wire without changing any other factors.
(10) Assuming both wires are initially at [tex]$T_0 = 20 \, \degree\text{C}$[/tex], we can calculate the final temperature T of the original wire when its resistance is equal to R, the resistance of the second wire.
Since the resistance of a wire depends on temperature, we can use the temperature coefficient of resistance to calculate the change in resistance.
Given:
- Temperature coefficient, a = 0.004579, 1°C
The change in resistance can be calculated using the formula:
[tex]\[ \Delta R = R_0 \cdot a \cdot \Delta T \][/tex]
where [tex]$\Delta R$[/tex] is the change in resistance, [tex]$R_0$[/tex] is the initial resistance, a is the temperature coefficient, and [tex]$\Delta T$[/tex] is the change in temperature.
To make the resistances of the original and second wires equal, [tex]$\Delta R$[/tex] should be equal to [tex]$R - R_0$[/tex]. Solving for [tex]$\Delta T$[/tex]:
[tex]$\[ \Delta T = \frac{{R - R_0}}{{R_0 \cdot a}} \][/tex]
Substituting the given values:
[tex]$\[ \Delta T = \frac{{0.0506 - 0.01159}}{{0.01159 \cdot 0.004579}} \][/tex]
Calculating the expression:
[tex]$\[ \Delta T \approx 687.6 \, \degree\text{C} \][/tex]
Adding [tex]$\Delta T$[/tex] to the initial temperature [tex]$T_0$[/tex]:
[tex]\[ T = T_0 + \Delta T \][/tex]
Substituting the given values:
[tex]\[ T \approx 20 + 687.6 \][/tex]
Calculating the expression:
[tex]\[ T \approx 707.6 \, \degree\text{C} \][/tex]
Therefore, the final temperature T of the original wire, when its resistance is equal to the resistance of the second wire, is approximately [tex]$707.6 \, \degree\text{C}$[/tex].
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A hole in the ground in the shape of an inverted cone is 18 meters deep and has radius at the top of 13 meters. This cone is filled to the top with sawdust. The density, rho, of the sawdust in the hole depends upon its depth, x : rho(x)=2.1−1.5e−1.5xkg/m3.
A hole in the ground in the shape of an inverted cone is 18 meters deep and has radius at the top of 13 meters. This cone is filled to the top with sawdust. The density, rho, of the sawdust in the hole depends upon its depth. The mass of the sawdust in the hole is 6689.707396545126 kg.
The density of the sawdust in the hole is given by rho(x)=2.1−1.5e−1.5xkg/m3. This function gives the density of the sawdust at a depth of x meters. The volume of the sawdust in the hole can be calculated using the formula for the volume of a cone:
V = (1/3)πr2h
In this case, r = 13 and h = 18, so the volume of the sawdust is V = 1540.5 m3. The mass of the sawdust is then given by V * rho(x), which is approximately 6689.707396545126 kg.
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If f(x) = e^1/x , thenf′(x) = _____
The derivative of f(x) = e^(1/x) is f'(x) = -e^(1/x) / x^2.
To find the derivative of f(x) = e^(1/x), we can use the chain rule. Let's denote g(x) = 1/x. The chain rule states that if we have a composite function f(g(x)), then the derivative of f with respect to x is given by f'(g(x)) * g'(x).
In this case, f(g(x)) = e ^g(x), where g(x) = 1/x. The derivative of g(x) with respect to x is g'(x) = -1/x^2. Now, we can find the derivative of f(g(x)) using the chain rule.
f'(g(x)) = e ^g(x) * g'(x) = e^(1/x) * (-1/x^2) = -e^(1/x) / x^2.
So, the derivative of f(x) = e^(1/x) is f'(x) = -e^(1/x) / x^2.
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3. Find the explicit solution for the given homogeneous DE (10 points) y! x2 + y2 ху
Given that the differential equation (DE) is y! x² + y² хуWe are required to find the explicit solution for the homogeneous DE. The solution for the homogeneous differential equation of the form
dy/dx = f(y/x) is given by the substitution y = vx. In our problem, the equation is y! x² + y² ху
To solve this equation, we substitute y = vx and differentiate with respect to x. y = vx Substitute the value of y into the given differential equation.
( vx )! x² + ( vx )² x (vx)
= 0x! v! x³ + v² x³
= 0
Factor out x³ from the above equation.
x³ (v! + v²) = 0x
= 0, (v! + v²) = 0
⇒ v! + v² = 0
Divide both sides by v², we have
(v!/v²) + 1 = 0
⇒ d(v/x)/dx + 1/x = 0
Now integrate both sides with respect to x.
d(v/x)/dx = - 1/xv/x
= - ln|x| + C1
where C1 is the constant of integration.Substitute the value of
v = y/x back into the above equation.
y/x = - ln|x| + C1 y
= - x ln|x| + C1x
Thus, the solution of the homogeneous differential equation is y = - x ln|x| + C1x.
Therefore, the explicit solution for the given homogeneous DE is y = - x ln|x| + C1x.
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You deposit $1000 at 2% per year. What is the balance at the end of one year if the interest paid is compounded monthly? Select one: $2020 $1020.18 $3000 $1020
Therefore, the balance at the end of one year if you deposit $1000 at 2% per year if the interest paid is compounded monthly is $1020.18.
If the interest paid is compounded monthly, the balance at the end of one year if you deposit $1000 at 2% per year would be $1020.18.
Interest is the amount of money that you have to pay when you borrow money from someone or a financial institution. It is the charge that the borrower has to pay for the privilege of using the lender's money over time.
Compounding interest implies that interest will be earned on both the principal amount and any interest received on the money over time.
A few times each year, the interest gets compounded with this kind of interest. Each time interest is compounded, the new balance earns interest. The process keeps repeating until the end of the loan or investment period.
In this case, the annual interest rate is 2%.
The interest rate, however, is compounded monthly, which means that the annual interest rate is split into 12 equal parts and applied to your account balance each month.
Therefore, the effective interest rate is 2%/12 or 0.16667%.The formula for calculating interest compounded monthly is given as
A = P(1 + r/n)^(nt)
Where,
A = the balance after t years
P = the principal amount
r = the annual interest rate
n = the number of times the interest is compounded each year
t = the time in years.
Since the investment is made for 1 year, the above equation becomes
A = 1000(1 + 0.02/12)^(12*1)
= $1,020.18
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The probability distribution of the random variable X is shown in the accompanying table: Find P(X≥0),P(−2≤X≤2) and P(X≤3).
P(X≥0)=0.37
P(−2≤X≤2)=0.57
P(X≤3)=1
P(X≥0)=0.34
P(−2≤X≤2)=0.57
P(X≤3)=1
P(X≥0)=0.44
P(−2≤X≤2)=0.58
P(X≤3)=1
P(X≥0)=0.34
P(−2≤X≤2)=0.59
P(X≤3)=1
The probability distribution of the random variable X is shown in the accompanying table, P(X≥0) = 0.37, P(−2≤X≤2) = 0.57, P(X≤3) = 1.
We need to find the following probabilities: P(X≥0), P(−2≤X≤2), and P(X≤3).
The given table represents a discrete probability distribution, since the sum of the probabilities is 1.
In order to find P(X≥0), we need to add all probabilities that are equal to or greater than 0.
By looking at the table, we can see that only one probability value is given that is greater than or equal to 0: P(X=0) = 0.37.
Therefore, P(X≥0) = 0.37.To find P(−2≤X≤2), we need to add all probabilities that fall between -2 and 2 inclusive.
From the table, we can see that three probability values satisfy this condition:
P(X=-1) = 0.09, P(X=0) = 0.37, and P(X=1) = 0.11.
Therefore, P(−2≤X≤2) = 0.09 + 0.37 + 0.11 = 0.57.
To find P(X≤3), we need to add all probabilities that are less than or equal to 3.
From the table, we can see that all probabilities satisfy this condition: P(X=-1) = 0.09, P(X=0) = 0.37, P(X=1) = 0.11, P(X=2) = 0.06, and P(X=3) = 0.37.
Therefore, P(X≤3) = 0.09 + 0.37 + 0.11 + 0.06 + 0.37 = 1.
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Determine the global extreme values of the function f(x,y)=4x3+4x2y+5y2 ,x,y≥0,x+y≤1
fmin = ___
fmax = ___
Note: You can earn partial credit on this problem.
The actual minimum value is approximately -2.859 and occurs at the point (15/4, -3/2), while the actual maximum value is 2749 and occurs at the point (7, 10).
To find the global extreme values of the function f(x, y) = 4x³ + 4x²y + 5y², subject to the constraints x, y ≥ 0 and x + y ≤ 1, we need to consider the critical points in the interior of the region and on the boundary.
Step 1: Critical points in the interior of the region
To find the critical points, we need to take the partial derivatives of f(x, y) with respect to x and y and set them equal to 0.
∂f/∂x = 12x² + 8xy
∂f/∂y = 4x² + 10y
Setting ∂f/∂x = 0:
12x² + 8xy = 0
4x(3x + 2y) = 0
This gives two possibilities:
x = 0
3x + 2y = 0 --> y = -(3/2)x
Setting ∂f/∂y = 0:
4x² + 10y = 0
10y = -4x²
y = -(2/5)x²
So, the critical points in the interior are (0, 0) and (x, -(2/5)x²) where x can vary.
Step 2: Critical points on the boundary
Now, we need to consider the boundary of the region x + y ≤ 1.
Case 1: x = 0
In this case, we are restricted to y ≤ 1, so the critical point is (0, y) where 0 ≤ y ≤ 1.
Case 2: y = 0
In this case, we are restricted to x ≤ 1, so the critical point is (x, 0) where 0 ≤ x ≤ 1.
Case 3: x + y = 1
Substituting x + y = 1 into f(x, y), we get:
f(x, 1 - x) = 4x³ + 4x²(1 - x) + 5(1 - x)²
Simplifying, we have:
f(x, 1 - x) = 4x³ + 4x² - 4x³ + 5(1 - 2x + x²)
f(x, 1 - x) = 5x² - 10x + 5
Now, we need to find the extreme values of f(x, y) at the critical points.
Evaluate f(x, y) at the critical points:
f(0, 0) = 0
f(x, -(2/5)x²) = 4x³ + 4x²(-(2/5)x²) + 5(-(2/5)x²)²
f(x, -(2/5)x²) = 4x³ - (8/5)x⁴ + (2/5)x⁴
f(x, -(2/5)x²) = 4x³ - (6/5)x⁴
f(x, 1 - x) = 5x² - 10x + 5
Now, we can compare the values of f(x, y) at these critical points to find the minimum and maximum values.
Minimum value (fmin):
fmin = min{f(0, 0), f(x, -(2/5)x²), f(x, 1 - x)}
Maximum value (fmax):
fmax = max{f(0, 0), f(x, -(2/5)x²), f(x, 1 - x)}
Critical points:
To find the critical points, we need to determine where the gradient of f(x, y) is equal to zero.
The gradient of f(x, y) is given by:
∇f(x, y) = (12x² + 8xy, 4x² + 10y)
Setting each component of the gradient equal to zero, we get:
12x² + 8xy = 0 ...(1)
4x² + 10y = 0 ...(2)
From equation (2), we can solve for y in terms of x:
y = -4x²/10
y = -2x²/5 ...(3)
Substituting equation (3) into equation (1), we get:
12x² + 8x(-2x²/5) = 0
12x² - 16x³/5 = 0
60x² - 16x³ = 0
4x²(15 - 4x) = 0
This equation has two solutions: x = 0 and x = 15/4.
For x = 0, using equation (3) we find y = 0.
For x = 15/4, using equation (3) we find y = -2(15/4)²/5 = -3/2.
Therefore, the critical points are (0, 0) and (15/4, -3/2).
Endpoints of the region:
The endpoints of the region are (0, 0), (7, 0), and (7, 10).
Now we evaluate the function at the critical points and endpoints:
f(0, 0) = 4(0)³ + 4(0)²(0) + 5(0)² = 0
f(15/4, -3/2) = 4(15/4)³ + 4(15/4)²(-3/2) + 5(-3/2)² ≈ -2.859
f(7, 0) = 4(7)³ + 4(7)²(0) + 5(0)² = 1372
f(7, 10) = 4(7)³ + 4(7)²(10) + 5(10)² = 2749
Comparing these values, we find:
Minimum value (fmin):
fmin = -2.859 at (15/4, -3/2)
Maximum value (fmax):
fmax = 2749 at (7, 10)
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find a formula for an for the arithmetic sequence:a1=-1,a5=7
Answer:
[tex]a_{n}[/tex] = 2n - 3
Step-by-step explanation:
the nth term of an arithmetic sequence is
[tex]a_{n}[/tex] = a₁ + d(n - 1)
where a₁ is the first term and d the common difference
given a₁ = - 1 and a₅ = 7 , then
a₁ + 4d = 7 , that is
- 1 + 4d = 7 ( add 1 to both sides )
4d = 8 ( divide both sides by 4 )
d = 2
then
[tex]a_{n}[/tex] = - 1 + 2(n - 1) = - 1 + 2n - 2 = 2n - 3
[tex]a_{n}[/tex] = 2n - 3
Let θ^θ^ and θ~θ~ be two alternative unbiased estimators for the unknown parameter θθ. θ^θ^ is said to be (the most) efficient only if
a.E(θ^)=0E(θ^)=0.
b.var(θ^)
c.E(θ^)=θE(θ^)=θ.
d.var(θ^)var(θ^) is the minimum within the group of all linear unbiased estimators for θθ.
θ^ is said to be (the most) efficient only if var(θ^) is the minimum within the group of all linear unbiased estimators for θθ. Therefore, the option d.
Given that θ^ and θ~ be two alternative unbiased estimators for the unknown parameter θθ. var(θ^) is the minimum within the group of all linear unbiased estimators for θθ. The efficiency of an estimator is measured by its variance. An efficient estimator is an estimator that attains the lowest possible variance. This is obtained by the Cramér-Rao lower bound, which states that the variance of an estimator is bounded by the reciprocal of the Fisher information. In other words, the more information in the data, the more efficient the estimator is. Moreover, in the case of unbiased estimators, the one with the smallest variance is said to be the most efficient. Furthermore, an estimator is considered the most efficient if and only if its variance is equal to the Cramér-Rao lower bound.To know more about linear unbiased estimators, visit:
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A cylinder of radius r and height h has volume given by V=∏r
2
h. Find the volume of a cylindrical tin can of radius 8 cm and height 21.2 cm. Group 7
The volume of the cylindrical tin can is approximately 4288.65 cubic centimeters.
To find the volume of a cylindrical tin can, we can use the formula V = π[tex]r^2[/tex]h, where V represents the volume, r is the radius, and h is the height of the cylinder. In this case, the given radius is 8 cm and the height is 21.2 cm.
Calculate the base area
The base area of the cylinder can be found using the formula A = π[tex]r^2[/tex]. Plugging in the given radius, we have A = π[tex](8 cm)^2[/tex]. Simplifying this, we get A = 64π [tex]cm^2[/tex].
Multiply the base area by the height
Next, we multiply the base area by the height of the cylinder. Multiplying 64π [tex]cm^2[/tex] by 21.2 cm gives us the volume V = 1356.8π [tex]cm^3[/tex].
Approximate the value of π and calculate the volume
To find the approximate value of the volume, we substitute the value of π as 3.14. Multiplying 1356.8π [tex]cm^3[/tex] by 3.14, we get V ≈ 4269.632[tex]cm^3[/tex].
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Express the sum 21.1956+16.348+31.02+14.0 using the correct number of significant figures
82.56360
82.5636
82.564
82.56
82.6
83
83.0
Rounding this to one decimal place, the answer is 82.6. Thus, the correct expression of the sum with the appropriate number of significant figures is 82.6.
To determine the correct number of significant figures in the sum, we need to consider the rules for significant figures during addition. The rule states that the sum or difference of numbers should have the same number of decimal places as the number with the fewest decimal places.
In the given numbers, the number with the fewest decimal places is 14.0, which has one decimal place. Therefore, the sum should be rounded to one decimal place.
Calculating the sum, we get 21.1956 + 16.348 + 31.02 + 14.0 = 82.5636.
Rounding this to one decimal place, the answer is 82.6. Thus, the correct expression of the sum with the appropriate number of significant figures is 82.6.
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3. A lecturer takes a bag of chocolates to each lecture.At one lecture, her bag contains exactly 12 chocolates and she decides that she will ask 12 revision questions at this lecture. She estimates that for each question, there is a 90% chance that the first person to answer the question will get it correct and receive one chocolate. Let X be the number of chocolates that she gives out in the lecture. (Assume that chocolates are only given out when the first person to answer a question gets the question correct.)
(b) At the next lecture, she realises she only has four chocolates left in her bag. She decides to ask harder questions. She estimates that for each question there is 70% chance a student answers it correctly. Let H be the number of incorrect answers the lecturer has received before getting three correct answers from students and thus has given away all her chocolates. (Note: We are not concerned about how many questions have been asked, just the number of incorrect answers.)
i. Name the distribution (including its parameter(s)) that could be used to model H. State any assumptions you are making in using this model.
ii. Write down the probability mass function, fi (h), of H.
(b)
i. The distribution that could be used to model H is the negative binomial distribution. The negative binomial distribution models the number of failures before a specified number of successes occur. In this case, the number of incorrect answers (failures) before three correct answers (successes) are obtained.
Assumptions:
Each question is independent of others, and the probability of a student answering a question correctly remains constant.
The lecturer has an unlimited supply of questions to ask.
ii. The probability mass function (PMF) of the negative binomial distribution is given by:
fi(h) = C(h + r - 1, h) * p^r * (1 - p)^h
Where:
fi(h) represents the probability mass function of H for a given value of h (number of incorrect answers).
C(h + r - 1, h) represents the combination formula, which calculates the number of ways to choose h failures before obtaining r successes.
p is the probability of a student answering a question correctly.
r is the number of successes needed (in this case, 3 correct answers).
In this case, the PMF of H can be written as:
fi(h) = C(h + 3 - 1, h) * 0.7^3 * (1 - 0.7)^h
The negative binomial distribution with parameters r = 3 and p = 0.7 can be used to model H, the number of incorrect answers the lecturer receives before getting three correct answers and giving away all her chocolates.
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Sample size calculations 10 MARKS (a) A researcher wants to estimate the mean daily sugar intake among the 1,000 adults in their local town. They decide to take a random sample. In a small pilot study, the mean daily sugar intake from all sources was 36 grams and the standard deviation was 6 grams. How large a sample of adults should be taken if they want the margin of error of their estimated mean to be no larger than 1 gram? Did the finite population correction adjustment make much difference? Comment on why you think it did or it didn't. (5 MARKS) n
0
=
d
2
z
2
s
2
,n=
1+
N
n
0
n
0
] Note: Use z=1.96 (b) The same researcher wants to estimate the prevalence of diabetes in the same town. In a similar town it was estimated that 10% of adults have diabetes. The researcher wants to determine the percentage of adults have diabetes in their town by taking a simple random sample. How large should this sample be if the margin of error of the estimate is to be no larger than 2 percentage points (0.02) ? Did the finite population correction adjustment make much difference? Comment on why you think it did or it didn't. (5MARKS) n
0
=
d
2
z
2
p
(1−
p
)
,n=
1+
N
(n
0
−1)
n
0
Note: use z=1.96
(a) To estimate the mean daily sugar intake with a margin of error no larger than 1 gram, the researcher needs a sample size of 97 adults. The finite population correction adjustment did not make much difference because the sample size is relatively small compared to the population size.
(b) To estimate the prevalence of diabetes with a margin of error no larger than 2 percentage points, the researcher needs a sample size of 384 adults. The finite population correction adjustment did not make much difference because the population size is large and the sample size is relatively small.
(a) The formula to calculate the sample size for estimating the mean is given as n0 = (d^2 * z^2 * s^2) / [(d^2 * z^2 * s^2) + N], where d is the desired margin of error, z is the z-score corresponding to the desired level of confidence (1.96 for a 95% confidence interval), s is the standard deviation of the pilot study, and N is the population size. Plugging in the given values, we find n0 = 97. The finite population correction adjustment did not make much difference because the population size (1,000) is much larger than the sample size.
(b) The formula to calculate the sample size for estimating the prevalence is given as n0 = (d^2 * z^2 * p * (1-p)) / [(d^2 * z^2 * p * (1-p)) + (N * (n0-1))], where p is the estimated prevalence, and all other variables have the same meanings as in part (a). Plugging in the given values, we find n0 = 384. The finite population correction adjustment did not make much difference because the population size is large (not specified) and the sample size is relatively small.
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Convert 4.532×104 square feet to m2. 1.38×104 m2 4.210×103 m2 1.381×102 m2 4.878×103 m2 4.21×103 m2 Fone of the other answers provided is correct 488×103 m2 4210 m2 Choose all correct answers. If you have two quantities. A and B, with different units, which of the following operations are allowed? imagine, for example, that AB in m and B is is s, or thay A is in kg and B is in ∘C (degrees Celsiusl. −8+A A+B BA AB A tanal A/B A2+A2
The correct conversion of 4.532×10^4 square feet to m^2 is 4.210×10^3 m^2. The allowed operations when dealing with quantities of different units are A+B, A-B, A*B, A/B, and A^2.
To convert square feet to square meters, we need to know the conversion factor. The conversion factor for area units between square feet and square meters is 1 square meter = 10.764 square feet.
Therefore, to convert 4.532×10^4 square feet to square meters, we divide the given value by the conversion factor:
4.532×10^4 square feet / 10.764 = 4.210×10^3 square meters.
Hence, the correct conversion is 4.210×10^3 m^2.
Regarding the operations with quantities of different units, the following operations are allowed:
Addition (A + B) when both A and B have the same units.
Subtraction (A - B) when both A and B have the same units.
Multiplication (A * B) to combine different units (e.g., m * s).
Division (A / B) to divide different units (e.g., m / s).
Squaring (A^2) to calculate the square of a quantity with units.
Thus, A + B, A - B, A * B, A / B, and A^2 are all allowed operations.
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Capital Accumulation as a Source of Growth - Work It Out Question Country A and country B both have the production function. Y=F(K,L)=K
1/2
L
1/2
c. Assume that neither country experiences population growth or technological progress and that 6 percent of capital depreciates each year. Assume further that country A saves 10 percent of output each year and country B saves 16 percent of output each year. Using your answer from part b and the steady-state condition that investment equals depreciation, find the steady-state level of capital per worker (k
∗
), income per worker (y
∗
), and consumption per worker ( c
∗
) for each country. For Country A For Country B k
∗
for Country A: k
∗
for Country B: y
∗
for Country A: y
∗
for Country B: c
∗
for Country A: c
∗
for Country B:
The production function of both country A and B. It's assumed that neither country experiences population growth or technological progress and that 6 percent of capital depreciates each year.
It's further assumed that country A saves 10 percent of output each year and country B saves 16 percent of output each year.Using the steady-state condition that investment equals depreciation, the steady-state level of capital per worker (k*), income per worker (y*), and consumption per worker (c*) for each country are calculated.
The formula for steady-state output per worker is y* = (sF(k*) - δk*) / L where s is the savings rate, δ is the depreciation rate, and L is the labor force size. For Country A Steady-state investment per worker will b Steady-state consumption per worker Steady-state output per worker For Country B: Steady-state investment per worker consumption per worker
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4. Let X, Y, Z have joint pdf fx.v.z®,M+ z) =k(2+y+z) for 0≤x≤ 1, 0≤ y ≤1, 0≤ z≤1. (a) Find k. (b) Find fx(xly, z) and fz(zlx,y).
The value of integration ∫∫∫fx.v.z(x,y,z)dxdydz = 1∴ k/3 = 1 ⇒ k = 3
Given, the joint pdf of three random variables X, Y, and Z is given by: fx.v.z(x,y,z) = k(2+y+z) 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1
(a) To find k, we need to integrate the joint pdf over the entire range of the random variables: ∫
∫∫fx.v.z(x,y,z)dxdydz = 1
∫∫∫k(2+y+z)dxdydz = 1 [0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
∫∫k(2+y+z)dx[0 ≤ x ≤ 1]
∫k[x(2+y+z)]dy[0 ≤ y ≤ 1]
k[x(2+y+z)y]z[0 ≤ z ≤ 1]
∫∫kx(2+y+z)dydz[0 ≤ x ≤ 1]
∫kx[y(2+z)+yz]dz[0 ≤ y ≤ 1]
kx[yz + (2+z)/2]z[0 ≤ z ≤ 1]
kx[yz^2/2 + z^2/2 + z(2+z)/2][0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
Integrating w.r.t z: kx[y(z^3/3+z^2/2+(2/2)z)][0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
Substituting the limits of integration:
k/3 [0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
k/3 ∫∫[0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
Therefore, ∫∫∫fx.v.z(x,y,z)dxdydz = 1∴ k/3 = 1 ⇒ k = 3
(b) We need to find the marginal pdfs fx(x, y, z) and fz(z, x, y).
fx(x, y, z) = ∫f(x, y, z)dydz[0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
fx(x, y, z) = k ∫(2+y+z)dydz[0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
fx(x, y, z) = k [y(2+y+z)/2 + yz + z^2/2][0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
fx(x, y, z) = 3/2 [y(2+y+z)/2 + yz + z^2/2][0 ≤ y ≤ 1, 0 ≤ z ≤ 1]
fz(z, x, y) = ∫f(x, y, z)dxdy[0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
fz(z, x, y) = k ∫(2+y+z)dxdy[0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
fz(z, x, y) = k [(2+y+z)/2 x + (2+y+z)/2 y + xy][0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
fz(z, x, y) = 3/2 [(2+y+z)/2 x + (2+y+z)/2 y + xy][0 ≤ x ≤ 1, 0 ≤ y ≤ 1]
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PLEASE ANSWER DUE IN 9 MINS WILL GIVE BRAINLIEST!!
Answer:
176in^2
Step-by-step explanation:
Total Surface Area:
2*16+4*36=32+144=176in^2
Solve each equation on the interval 0≤θ≤2π.
a) cosθ= 1/2
b) cosθ=−√3/2
The solutions on interval cosθ= 1/2 would be π/3 and 5π/3, the solutions on interval cosθ=−√3/2 would be 5π/6 and 7π/6.
a) The given equation is cosθ = 1/2 on the interval 0≤θ≤2π.Therefore, we need to find the solution of the equation on the given interval.Let's draw the unit circle to determine the solution of the given equation.
We can see that when we draw a line at an angle of θ degrees from the positive x-axis, the point of intersection between the line and the unit circle is (cosθ, sinθ).Now we can see that the line will intersect with the unit circle at two points making two angles with the positive x-axis as shown below.Let the two angles be A and B then cos A = cos B = 1/2So A = π/3 or 2π/3 and B = 4π/3 or 5π/3We know that the interval 0 ≤ θ ≤ 2π. Therefore, the solutions on this interval are π/3 and 5π/3.
b) The given equation is cosθ=−√3/2 on the interval 0≤θ≤2π.Let's draw the unit circle to determine the solution of the given equation.We can see that when we draw a line at an angle of θ degrees from the positive x-axis, the point of intersection between the line and the unit circle is (cosθ, sinθ).Now we can see that the line will intersect with the unit circle at two points making two angles with the positive x-axis as shown below.Let the two angles be A and B then cos A = cos B = −√3/2So A = 5π/6 or 7π/6 and B = 11π/6 or π/6We know that the interval 0 ≤ θ ≤ 2π.
Therefore, the solutions on this interval are 5π/6 and 7π/6.
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A student is factoring a trinomial using 1-Grouping (also known as the AC-Method). Using your knowledge of the steps in this method, complete the student's work by filling in each of the boxes below (do not use any spaces when you type your answers). HINT: It may be helpful to fill in the boxes from the bottom to the top.
20x ^2−
=20x ^2+15x−18
=5x
=(4x+3)(5x−3)−6(4x+
The student is factoring a trinomial using the 1-Grouping (AC-Method) technique. The completed steps are as follows:
20x^2−15x−18
= 20x^2+15x−18 (rearranging the terms)
= 5x(4x+3)−6(4x+3) (grouping the terms)
= (4x+3)(5x−6) (factoring out the common binomial)
Explanation: To factor the trinomial using the 1-Grouping (AC-Method), the student needs to follow the steps correctly. Here's a breakdown of the completed steps:
1. Start with the trinomial 20x^2−15x−18.
2. Rearrange the terms to obtain 20x^2+15x−18.
3. Identify the terms that have a common factor, which in this case is 5x. Factor out 5x from the first two terms: 5x(4x+3).
4. Identify the terms that have a common factor, which is 6. Factor out 6 from the last two terms: −6(4x+3).
5. Now, the common binomial factor (4x+3) can be factored out, resulting in the final factored form: (4x+3)(5x−6).
By following these steps, the student successfully factored the trinomial using the 1-Grouping (AC-Method).
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Suppose that the time in minutes required to serve a customer at a certain facility has an exponential distribution for which the value of the parameter θ is unknown and that the prior distribution of θ is a gamma distribution for which the mean is 0.3 and the standard deviation is 1. If the average time required to serve a random sample of 40 customers is observed to be 9.6 minutes, what is the posterior distribution of θ?
The posterior distribution of θ is a gamma distribution with parameters 40 + 0.09 and 9.6 + 0.3Posterior(θ | X) ~ Gamma(40.09, 9.9)
To determine the posterior distribution of θ, we can use Bayes' theorem. Let's denote:
- X: Average time required to serve a random sample of 40 customers (9.6 minutes)
- θ: Parameter of the exponential distribution
- Prior distribution of θ: Gamma distribution with mean 0.3 and standard deviation 1
We can express the posterior distribution of θ as:
Posterior(θ | X) ∝ Likelihood(X | θ) * Prior(θ)
Given that the exponential distribution is characterized by the parameter θ, the likelihood function can be expressed as:
Likelihood(X | θ) = (1/θ)^n * exp(-X/θ)
Where n is the sample size (40 in this case).
The prior distribution of θ is given as a gamma distribution with mean 0.3 and standard deviation 1. We can denote the gamma distribution as Gamma(α, β), where α is the shape parameter and β is the rate parameter. To find the specific values of α and β, we need to use the mean and standard deviation of the gamma distribution:
Mean = α/β = 0.3
Standard deviation = sqrt(α)/β = 1
From these equations, we can solve for α and β:
α = (mean/standard deviation)^2 = (0.3/1)^2 = 0.09
β = mean/standard deviation^2 = 0.3/(1^2) = 0.3
Now, we can calculate the posterior distribution by multiplying the likelihood and the prior distribution:
Posterior(θ | X) ∝ (1/θ)^n * exp(-X/θ) * θ^(α-1) * exp(-βθ)
Simplifying the expression:
Posterior(θ | X) ∝ θ^(n + α - 1) * exp(-(X/θ + βθ))
We recognize this expression as the kernel of a gamma distribution. Therefore, the posterior distribution of θ is a gamma distribution with parameters n + α and X + β.
In this case,the posterior distribution of θ is a gamma distribution with parameters 40 + 0.09 and 9.6 + 0.3.
Posterior(θ | X) ~ Gamma(40.09, 9.9)
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Five gasoline stations are located in a region such that any one station is exactly 1 mile away from at least two other stations. This is shown in the diagram to the right. You are currently at station A but believe the following to be true about the distribution of price that could be charged by any other station (each price is equally likely Price/gal. Pe(price) 2.00 020 2.20 0.20 1.80 0.20 1.60 0.20 2.40 020 B 1 mile of the time and travel expense to visit another station 1 mile away is $0, what is the most you would be willing to pay for a gallon of gas at station A? The most you would be willing to pay for a gallon of gas at station Als $ 2. (round your answer to the nearest penny) Suppose you find out for certain that station Fin charging $18/gallon the distribution of prices for other stations is unchanged) The most you would be willing to pay for a gallon of gas at station Als $ (round your answer to the nearest periny)
Given, there are five gasoline stations located in a region such that any one station is exactly 1 mile away from at least two other stations. The diagram is shown below: Thus, we can see that the station A is 1 mile away from stations B and C.
We are currently at station A but believe the following to be true about the distribution of price that could be charged by any other station. (each price is equally likely Price/gal. Pe(price) 2.00 0.20 2.20 0.20 1.80 0.20 1.60 0.20 2.40 0.20) Let, the most you would be willing to pay for a gallon of gas at station A be x. Then, the cost of visiting stations B and C are 0 as they are 1 mile away from station A. Therefore, the average cost of a gallon of gas at station A, \frac{x + 2.20 + 1.80}{3} = \frac{x + 4.00}{3} As given, all prices are equally likely. So, the expected value is the sum of products of each possible price and its probability.
Hence, the expected cost of a gallon of gas at station A is:
Expected cost of a gallon of gas at station A = 2.00(0.2) + 2.20(0.2) + 1.80(0.2) + 1.60(0.2) + 2.40(0.2)
= $2.00
Now, we know that station F is charging $1.8 per gallon of gas. So, the expected cost of a gallon of gas at station A is: Expected cost of a gallon of gas at station A = 2.00(0.2) + 2.20(0.2) + 1.60(0.2) + 2.40(0.2)
= $2.00
Thus, the most you would be willing to pay for a gallon of gas at station A, given that station F is charging $1.8 per gallon of gas is $2.
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The rate of crude oil production from 2009 to 2011 by Pemex, Mexico's national oil company, can be approximated by q(t) = 6.2t^2 − 146t + 1,910 million barrels per year (9 ≤ t ≤ 11), where t is time in years since the start of 2000. During that time, Mexico exported crude oil to the United States at a rate of r(t) = −14t^2 + 292t − 1,100 million barrels per year (9 ≤ t ≤ 11). Compute the area between the two curves using the limits t=9 and t=11. (Round your answer to the nearest whole number).
The area between the two curves q(t) and r(t) over the interval [9, 11] is approximately 3,840 million barrels per year. This represents the difference in crude oil production and export.
To compute the area between the two curves, we need to find the definite integral of the difference between the two functions over the interval [9, 11].
The area can be calculated as follows:
Area = ∫[9,11] (q(t) - r(t)) dt
Substituting the given functions q(t) and r(t), we have:
Area = ∫[9,11] (6.2t^2 - 146t + 1,910 + 14t^2 - 292t + 1,100) dt
Simplifying, we get:
Area = ∫[9,11] (20t^2 - 438t + 3,010) dt
Evaluating the integral, we find:
Area = [((20/3)t^3 - 219t^2 + 3,010t)] [9,11]
Plugging in the upper and lower limits, we have:
Area = ((20/3)(11)^3 - 219(11)^2 + 3,010(11)) - ((20/3)(9)^3 - 219(9)^2 + 3,010(9))
Calculating the expression, the approximate area between the two curves is 3,840 million barrels per year.
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1) sample of voters were polled to determine the likelyhood of measure 324 passing. The poll determined that 76 % of voters were in favor of the measure with a margin of error of 2.2 %. Find the confidence interval. Use ( ) in your notation.
2) The mean was found to be 50% and the confidence interval was (48%,52%) therefore the margin of error was +/- _____%.
3)The confidence interval was (39%, 43%)
a. What was the margin of error? +/- %
b. What was the summary statistic? %
1) Confidence interval: (73.8%, 78.2%). (2). Margin of error was +/- 2%. (3) a) Margin of error was +/- 2%. b) The summary statistic was 41%.
(1) A sample of voters was polled to determine the likelihood of Measure 324 passing.
The poll determined that 76 % of voters were in favor of the measure with a margin of error of 2.2 %.
Find the confidence interval. Use ( ) in your notation.
The formula to find the confidence interval is given by:
Lower limit = Mean - Z (α/2) * σ / √n
Upper limit = Mean + Z (α/2) * σ / √n
Where:
Mean is the average, Z is the Z-value (e.g. 1.96 for a 95% confidence interval), σ is the standard deviation, and n is the sample size.
(2) The margin of error is calculated using the formula, margin of error = Z (α/2) * σ / √n.2) Margin of error was +/- 2%.
A confidence interval is an estimate of an unknown population parameter that provides a range of values that, with a certain degree of probability, contains the true value of the parameter.
The margin of error is a statistic that quantifies the range of values that we expect the true result to fall between when using a confidence interval. In this question, the mean was found to be 50% and the confidence interval was (48%,52%). We can deduce that the margin of error would be +/- 2% by calculating half of the difference between the upper and lower limits of the confidence interval. Thus, the margin of error, in this case, is 2%.
3) a) Margin of error was +/- 2%. b) The summary statistic was 41%.
A confidence interval is an estimate of an unknown population parameter that provides a range of values that, with a certain degree of probability, contains the true value of the parameter. In this question, the confidence interval was (39%, 43%). We can calculate the margin of error to be +/- 2% by taking half of the difference between the upper and lower limits of the confidence interval. Therefore, the margin of error is 2%. The summary statistic can be obtained by calculating the average of the upper and lower limits of the confidence interval. Thus, the summary statistic, in this case, is (39%+43%)/2 = 41%.
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a sample of a population taken at one particular point in time is categorized as:
Answer:
Cross-Sectional Study
Step-by-step explanation:
Want the correct answer?
I i The probability of obtaining a 7 is 1/5.
ii The probability of obtaining an odd number is 3/5.
2 i The probability of obtaining an odd sum is 13/25.
b The probability of obtaining a sum of 14 or more is 6/25.
c. The probability of obtaining the same number on all three spins is 1/125.
How to calculate the probabilityI(i) The probability of obtaining a 7 is 1 out of 5 since there is only one favorable outcome (spinning the number 7), and there are five possible outcomes (numbers 1, 3, 5, 7, and 9).
Therefore, the probability of obtaining a 7 is 1/5.
(ii) There are three favorable outcomes (numbers 1, 3, and 7) out of five possible outcomes.
Therefore, the probability of obtaining an odd number is 3/5.
(b) (a) Odd sum: Out of the 25 possible outcomes (5 numbers on the first spin multiplied by 5 numbers on the second spin), there are 13 combinations that result in an odd sum: (1, 1), (1, 3), (1, 5), (1, 7), (1, 9), (3, 1), (3, 3), (3, 5), (3, 7), (3, 9), (7, 1), (7, 3), (9, 1). Therefore, the probability of obtaining an odd sum is 13/25.
(b) Sum of 14 or more: There are six combinations that result in a sum of 14 or more: (7, 7), (7, 9), (9, 7), (9, 9), (7, 5), (5, 7). Therefore, the probability of obtaining a sum of 14 or more is 6/25.
(c) The probability of obtaining the same number on the first two spins is 1/5, and the probability of obtaining the same number on the third spin is also 1/5.
Therefore, the probability of obtaining the same number on all three spins is (1/5) * (1/5) * (1/5)
= 1/125.
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Answer the following questions about the Standard Normal Curve: a.) Find the area under the Standard Normal curve to the left of z=1.24 b.) Find the area under the Standard Normal curve to the right of z=−2.13 c.) Find the z-value that has 87.7% of the total area under the Standard Normal curve lying to the left of it. d.) Find the z-value that has 20.9% of the total area under the Standard Normal curve lying to the right of it.
a) The area under the standard normal curve to the left of z = 1.24 is 0.8925.
b) The area under the standard normal curve to the right of z = −2.13 is 0.9834
c) The z-score that has 87.7% of the total area under the standard normal curve lying to the left of it is 1.18.
d) The z-score that has 20.9% of the total area under the standard normal curve lying to the right of it is -0.82.
a.) Find the area under the Standard Normal curve to the left of z=1.24:
Using the z-table, the value of the cumulative area to the left of z = 1.24 is 0.8925
b.) Find the area under the Standard Normal curve to the right of z=−2.13:
Using the z-table, the value of the cumulative area to the left of z = −2.13 is 0.0166.
c.) Find the z-value that has 87.7% of the total area under the Standard Normal curve lying to the left of it:
Using the z-table, the closest cumulative area to 0.877 is 0.8770. The z-score corresponding to this cumulative area is 1.18.
d.) Find the z-value that has 20.9% of the total area under the Standard Normal curve lying to the right of it:
Using the z-table, the cumulative area to the left of z is 1 - 0.209 = 0.791. The z-score corresponding to this cumulative area is 0.82.
Note: The cumulative area to the right of z = -0.82 is 0.209.
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Required information A ball is thrown upward, from the ground, with an initial velocity of 17 m/s. The approximate value of g=10 m/s
2
. Take the upward direction to be positive. Tossed Ball Velocity up is positive v
a
=+20 m/sa=−10 m/s
2
down is negative v=v
0
+at NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the magnitude and the direction of the ball's velocity 2 seconds after it is thrown? The magnitude of the velocity is m/s, and the motion is
The magnitude of velocity of ball after 2 seconds of being thrown is 37 m/s.
Given values are:
Initial Velocity, u = 17 m/s
Acceleration due to gravity, g = 10 m/s²
Time, t = 2 s
The velocity of the ball at time t, v is given by
v = u + gt
Here, u = 17 m/s, g = 10 m/s², and t = 2 s
Putting the values, we get
v = u + gt
= 17 + 10 × 2
v = 17 + 20
v = 37 m/s
This velocity is positive since the ball is going upwards.
Therefore, the direction of the ball's velocity after 2 seconds of being thrown is upward, or positive.
The magnitude of velocity of ball after 2 seconds of being thrown is 37 m/s.
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A six-sided dice is rolled. Four points is scored if the roll comes up with 5 or 6 , one point if it comes up 1,2,3 or 4 . If x is the point reward, what is the variance of X. Give you answer-in the form a bc
The variance of X is -34/27.
The random variable X has two possible outcomes:
1 with probability of 2/6 = 1/3 or 4 with probability of 4/6 = 2/3.So, the expected value of X is:
E(X) = 1(1/3) + 4(2/3) = 3(2/3) = 11/3.
The squared deviation from the mean of a random variable is referred to as variance in probability theory and statistics. The square of the standard deviation is another common way to express variation. Variance is a measure of dispersion, or how far apart from the mean a group of data are from one another.
Now we can compute the variance of X by using the following formula:
Var(X) = E(X²) - [E(X)]².
The expected value of X² is:E(X²) = 1²(1/3) + 4²(2/3) = 29/3.
So,Var(X) = E(X²) - [E(X)]²= 29/3 - (11/3)²= 29/3 - 121/9= (87 - 121) / 27= - 34 / 27.
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