Scores on a test are normally distributed with a mean of 63 and a standard deviation of 9.3. Find P81, which separates the bottom 81% from the top 19%.

Answers

Answer 1

We want to find [tex]x[/tex] such that

[tex]\mathrm{Pr}(X \le x) = 0.81[/tex]

where [tex]X[/tex] is the random variable for test scores, and [tex]X\sim\mathrm{Normal}(63,9.3^2)[/tex].

Transform [tex]X[/tex] to [tex]Z\sim\mathrm{Normal}(0,1)[/tex] using the relation

[tex]X = \mu + \sigma Z \implies Z = \drac{X-63}{9.3}[/tex]

so that

[tex]\mathrm{Pr}(X \le x) = \mathrm{Pr}\left(\dfrac{X-63}{9.3} \le \dfrac{x-63}{9.3}\right) = \mathrm{Pr}\left(Z \le \dfrac{x-63}{9.3}\right) = 0.81[/tex]

Applying the inverse CDF of [tex]Z[/tex] (denoted by [tex]\Phi^{-1}[/tex]), we have

[tex]\dfrac{x-63}{9.3} = \Phi^{-1}(0.81) \approx 0.8779[/tex]

Solve for [tex]x[/tex].

[tex]\dfrac{x-63}{9.3} \approx 0.8779[/tex]

[tex]x-63 \approx 8.1644[/tex]

[tex]x \approx 76.1644[/tex]

Then the cutoff test score for the 81% percentile is [tex]P_{81} \approx \boxed{76}[/tex].


Related Questions

A bank ATM system has a pad with 10 digits (0 to 9). Find the number of possible 4-digit pin codes

if digits can be repeated.
if digits cannot be repeated.

a.
1. 10 000 ; 2. 5 040.

b.
1. 5 040; 2. 10 080.

c.
1. 10 000; 2. 210.

d.
1. 3 125; 2. 15 120.

Answers

Answer:

A

Step-by-step explanation:

If digits can be repeated, that means there are 10 options for each place in the pin code. 10*10*10*10 = 10,000

If digits can not be repeated, there are 10 options for the first digit, 9 options for the second digit, 8 options for the third digit, and 7 options for the fourth digit. 10*9*8*7 = 5040

domain of f(x)=(1/4)^x

What is the domain of f(x)

O A. x>0

OB. All real numbers

O C. y>0

O D. x<0

? Need help asap

Answers

Answer: B. All real numbers

Step-by-step explanation:

See attached image.

A textbook store sold a combined total of 204 math and psychology textbooks in a week. The number of math textbooks sold was three times the number of psychology textbooks sold. How many textbooks of each type were sold?​

Answers

51 psychology textbooks
153 math textbooks

Let’s let x equal the number of psychology textbooks sold. That means that the number of math textbooks sold would be 3x (3 times x, or the number of psychology textbooks). Knowing the total is 204, we can make an equation that says x (the number of psychology textbooks) plus 3x (the number of math textbooks) equals 204 (the total number of textbooks). We can then combine x plus 3x to make 4x. So we’re left with 4x equals 204. If we divide both sides of the equation by 4 we’re left with x equals 51. So 51 psychology textbooks were sold. We then multiply this number by three to get 153, the number of math textbooks sold.

Let x = number of psychology textbooks
So 3x = number of math textbooks
x + 3x = 204
4x = 204
x = 51 3x = 3(51) = 153

A total of 204 textbooks were sold with 153 math textbooks and 51 psychology textbooks being sold, as the number of math textbooks was three times the number of psychology textbooks.

Given that,

The combined total of math and psychology textbooks sold in a week is 204.

The number of math textbooks sold is three times the number of psychology textbooks sold.

To solve this problem,

Assign variables to represent the number of math and psychology textbooks sold.

Let's say "M" represents the number of math textbooks and "P" represents the number of psychology textbooks.

We know that the combined total of math and psychology textbooks sold is 204,

So the equation be:

M + P = 204  .....(i)

We are also given that the number of math textbooks sold was three times the number of psychology textbooks sold.

In equation form, this can be expressed as:

M = 3P

Now, we can substitute the value of M in terms of P into the equation (i):

3P + P = 204

Combining like terms, we get:

4P = 204

Dividing both sides by 4, we find:

P = 51

So, the number of psychology textbooks sold is 51.

To find the number of math textbooks sold,

Substitute this value back into the equation:

M = 3P

M = 3(51)

M = 153

Therefore, the number of math textbooks sold is 153.

Hence,

153 math textbooks and 51 psychology textbooks were sold.

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Which equation represents a parabola with a vertex at (7,-3)

Answers

Answer:

See Below

Step-by-step explanation:

y = a (x-7)^2 -3       where 'a'  is any number except 0

Which expression is equivalent to sec²x - 1?
O A. cot²x
OB. tan²x
OC. CsC²x
OD. cos²x

Answers

[tex]l = sec {}^{2} x - 1 \\ l = \frac{1}{cos {}^{2} x} - \frac{cos {}^{2} x}{cos {}^{2} x} \\ l = \frac{1 - cos {}^{2} x}{cos {}^{2}x } \\ l = \frac{sin {}^{2} x}{cos {}^{2} x} = ( \frac{sinx}{cosx} ) {}^{2} = tan {}^{2} x[/tex]

B

List the sides of triangle RST in acsending order (Shortest to longest.)

Answers

The sequence of angles from smallest to largest is m∠T,m∠S,m∠R.

Given three angles be m∠T=4x-52°,m∠S=x+38°,m∠R=2x+47°.

We are required to arrange the angles in ascending order.

Ascending order means values that are written in a way that smallest values come first and larger values comes last.

First we have to find the angles at one value of x.

The values will differ as the values of x differ so we have to choose point or value of x.

Suppose x=20.

Then the angles are:

m∠T=4x-52°

=4*20-52

=80-52

=28°

m∠S=x+38°

=20+38

=58°

m∠R=2x+47°

=2*20+47

=40+47

=87°

Hence the sequence of angles from smallest to largest is m∠T,m∠S,m∠R.

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what is ordinary numbers

Answers

What is ordinary number?

1 : a number designating the place (such as first, second, or third) occupied by an item in an ordered sequence — see Table of Numbers. 2 : a number assigned to an ordered set that designates both the order of its elements and its cardinal number.

Suppose you know that the distribution of sample proportions of fifth grade students in a large school district who read below grade level in samples of 100 students is normal with a mean of 0.30 and a standard deviation of 0.12. You select a sample of 100 fifth grade students from this district and find that the proportion who read below grade level in the sample is 0.54. This sample proportion lies 2.0 standard deviations above the mean of the sampling distribution. What is the probability that a second sample would be selected with a proportion greater than 0.54 ?

Answers

Based on the mean of the sample and the proportion who read below grade level, the probability that a second sample would have a proportion greater than 0.54 is 0.9772.

What is the probability of the second sample being greater than 0.54?

The probability that the second sample would be selected with a proportion greater than 0.54 can be found as:

P (x > 0.54) = P ( z > (0.54 - 0.30) / 0.12))

Solving gives:

P (x > 0.54) = P (z > 2)

P (x > 0.54) = 0.9772

In conclusion, the probability that the second sample would be selected with a proportion greater than 0.54 is 0.9772.

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The mean of a normally distributed data set is 110, and the standard deviation is 15.

a) Use the standard normal table to find the probability that a randomly-selected data value is greater than 95.

b) Use the standard normal table to find the probability that a randomly-selected data value is greater than 125.

Answers

From the standard normal table, the respective probabilities are; 0.908 and 0.841

How to find the probability from z-score?

Formula for calculating the standard score or z score is:

z = (x - μ)/σ

where:

z is the standard score

x is the raw score

μ is the population mean

σ is the population standard deviation

A) We are given;

x = 95

μ = 110

σ = 15

Thus;

z = (95 - 110)/15

z = 1.33

From the normal standard distribution table we can find the probability from the z-score as;

P(x > 95) = 1 - 0.091759.

P(x > 95) = 0.908

B) We are given;

x = 125

μ = 110

σ = 15

Thus;

z = (125 - 110)/15

z = 1

From the normal standard distribution table we can find the probability from the z-score as;

P(x > 95) = 1 - 0.158655.

P(x > 95) = 0.841

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Help me with this equation, please. (Image Attached)

Answers

So, the equation is sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)

The question has to to with trigonometric identities?

What are trigonometric identities?

Trigonometric identities are equations that show the relationship between the trigonometric ratios.

How to solve the equation?

Given the equation sin(x + y)/sin(x - y)

Using the trigonometric identities.

sin(x + y) = sinxcosy + cosxsiny andsin(x - y) = sinxcosy - cosxsiny

So, sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/(sinxcosy + cosxsiny)

Dividing the rnumerator and denominator of ight hand side by sinx, we have

sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/sinx/(sinxcosy + cosxsiny)/sinx

sin(x + y)/sin(x - y) = (sinxcosy/sinx + cosxsiny/sinx)/(sinxcosy/sinx + cosxsiny/sinx)

= (cosy + cotxsiny)/(cosy + cotxsiny) (since cosx/sinx = cotx)

Dividing the numerator and denominator of the right hand side by cosy, we have

= (cosy + cotxsiny)/cosy/(cosy + cotxsiny)/cosy

= (cosy/cosy + cotxsiny/cosy)/(cosy/cosy + cotxsiny/cosy)

= (1 + cotxstany)/(1 + cotxtany)   [since siny/cosy = tany]

So,  sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)

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Find the mean for the amounts: $17,484, $14,978, $13,521, $14,500, $18,540, $14,978

Answers

Answer:

$15666.83 (2dp)

Step-by-step explanation:

Mean = Total of all values / Number of Values

= [tex]\frac{17484+14978+13521+14500+18540+14978}{6}[/tex]

=[tex]\frac{94001}{6}[/tex]

= $15666.83 (2dp)

What’s an Inequality?

Answers

Answer:

mathematics, an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions. It is used most often to compare two numbers on the number line by their size.

Answer:

The quality of being unequal or uneven is known as the inequality.

Suppose f(x) = x2. What is the graph of g(x) = f(4x)?

Answers

[tex]g(x)=f(4x)=(4x)^2 = 16x^2[/tex]

The graph is shown in the attached image.

Question 10 of 25
Which pair of functions are inverses of each other?

Answers

[tex]f(x) = \sqrt[3]{11x} \\ y = \sqrt[3]{11x} \\ x = \sqrt[3]{11y} \\ x {}^{3} = 11y \\ y = \frac{x {}^{3} }{11} [/tex]

Option A eliminated

[tex]f(x) = \frac{x}{7} + 10 \\ y = \frac{x}{7} + 10 \\ x = \frac{y}{7} + 10 \\ x - 10 = \frac{y}{7} \\ y = \frac{x - 10}{7} [/tex]

Option B eliminated

[tex]f(x) = \frac{7}{x} - 2 \\ y = \frac{7}{x} - 2 \\ x = \frac{7}{y} + 2 \\ x - 2 = \frac{7}{y} \\ y = \frac{7}{x - 2} [/tex]

Option C eliminated

By elimination it's DConfirmation:

[tex]f(x) = 9x - 6 \\ y = 9x - 6 \\ x = 9y - 6 \\ x + 6 = 9y \\ y = \frac{x + 6}{9} = g(x)[/tex]

Write a quadratic function fwhose zeros are 2 and 8.
f(x) = 0

Answers

Factored Form:
f(x) = (x - 2) (x - 8)

Standard Form:
f(x) = x^2 - 10x + 16

What is the solution to 3/4 a>-16?
O a>-21/3
O a<-21
O a> 21-1/
O a <21/13

Answers

[tex] \frac{3a}{4} > - 16 \\ 3a > - 64 \\ a > \frac{ - 64}{3} \\ a > 21 \frac{1}{3} [/tex]

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{Equation:}[/tex]

[tex]\mathsf{\dfrac{3}{4}a > -16}[/tex]

[tex]\huge\textbf{Simplifying it:}[/tex]

[tex]\mathsf{\dfrac{3}{4}a > -16}[/tex]

[tex]\mathsf{\dfrac{3}{4}a > - \dfrac{16}{1}}[/tex]

[tex]\huge\textbf{Divide \boxed{\dfrac{4}{3}} to both sides:}[/tex]

[tex]\mathsf{\dfrac{4}{3}\times\dfrac{3}{4}a > -16\times\dfrac{4}{3}}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\mathsf{a > \dfrac{4}{3}\times -16}[/tex]

[tex]\mathsf{a > - \dfrac{16}{1} \times\dfrac{4}{3}}[/tex]

[tex]\mathsf{a > \dfrac{-16\times4}{1\times3}}[/tex]

[tex]\mathsf{a > \dfrac{-64}{3}}[/tex]

[tex]\mathsf{a > -\dfrac{64}{3}}[/tex]

[tex]\mathsf{a > -21 \dfrac{1}{3}}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{- \frak{21\dfrac{1}{3}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Consider the equation
x/x − 1 = 6x + 1/x − 1
What is the LCD?
Multiply both sides of the equation by the LCD and rewrite the resulting quadratic equation in general form. _____=0
Solve the equation and check the solutions in the original equation. (Enter your answers as a comma-separated list.) x=________

Answers

The solution to the original equation is 1, 1/6

Solving equation

Equations are expressions separated by mathematical operations.

Given the equation below

x/x − 1 = 6x + 1/x − 1

From the given expression, the least common denominator is x -1

Multiply both sides by x-1 to have;

x = 6x(x-1) +1

Expand

x = 6x^2-6x + 1

Equate to zero

6x^2-6x-x + 1 = 0

6x^2-7x +1= 0

The resulting quadratic equation in general form is 6x^2-7x +1 = 0

Factorize

6x^2 -6x-x + 1 = 0

Group the result

6x(x-1)-1(x-1) = 0

(6x-1)(x-1) = 0

6x - 1 = 0 and x -1 = 0

x = 1 and 1/6

Hence the solution to the original equation is 1, 1/6

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The entire graph of the function g is shown in the figure below.
Write the domain and range of g using interval notation.

Answers

The domain of the function is [-4, 4) and the range of the function is [-5, 2)

How to determine the domain and the range of the function?

The domain

As a general rule, it should be noted that the domain of a function is the set of input values or independent values the function can take.

This means that the domain is the set of x values

From the graph, we have the following intervals on the x-axis

x = -4 (closed circle)

x =4 (open circle)

This means that the domain of the function is [-4, 4)

The range

As a general rule, it should be noted that the range of a function is the set of output values or dependent values the function can produce.

This means that the range is the set of y values

From the graph, we have the following intervals on the y-axis

y = -5 (closed circle)

y = 2 (open circle)

This means that the range of the function is [-5, 2)

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A SINGLE CARD IS DRAWN AT RANDOM FROM A STANDARD DECK OF 52 CARDS. FIND THE PROBABILITY OF DRAWING THE FOLLOWING CARDS. PLEASE REDUCE TO LOWEST TERMS.
A) A DIAMOND OR A 5 __________
B) A HEART AND A JACK __________
C) A JACK OR AN 8 __________
D) A HEART OR A SPADE __________
E) A RED AND FACE CARD __________
F) A RED CARD OR A QUEEN __________

Answers

Answer:

A. [tex]\frac{17}{52}[/tex]

B. [tex]\frac{17}{52}[/tex]

C. [tex]\frac{2}{13}[/tex]

Step-by-step explanation:

A.

There are 52/4 diamonds in the deck and 4 '5's in the dech of cards

52/4 = 13 + 4 = 17

Therefore, you have a  [tex]\frac{17}{52}[/tex] chance of drawing one of those cards.

B.

There are 13 hearts in the deck and 4 jacks. Therefore, your odds are the same : [tex]\frac{17}{52}[/tex]

C.

There are 4 jacks in a deck of cards and 4 '8's in a deck of cards

Therefore your probability is [tex]\frac{8}{52}[/tex] which simplifies to [tex]=\frac{2}{13}[/tex]

As per brainly guidelines I can only answer 3 questions in one answer

A) 13/52 + 4/52 - 1/52 = 16/52 = 4/13
B) 1/52
C) 4/52 + 4/52 = 8/52 = 2/13
D) 13/52 + 13/52 = 26/52 = 1/2
E) 6/52 = 3/26
F) 26/52 + 4/52 - 1/52 = 29/52

please help me with these calculus bc questions

Answers

4. Compute the derivative.

[tex]y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1[/tex]

Find when the gradient is 7.

[tex]4x - 1 = 7 \implies 4x = 8 \implies x = 2[/tex]

Evaluate [tex]y[/tex] at this point.

[tex]y = 2\cdot2^2-2-1 = 5[/tex]

The point we want is then (2, 5).

5. The curve crosses the [tex]x[/tex]-axis when [tex]y=0[/tex]. We have

[tex]y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4[/tex]

Compute the derivative.

[tex]y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}[/tex]

At the point we want, the gradient is

[tex]\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}[/tex]

6. The curve crosses the [tex]y[/tex]-axis when [tex]x=0[/tex]. Compute the derivative.

[tex]\dfrac{dy}{dx} = 3x^2 - 4x + 5[/tex]

When [tex]x=0[/tex], the gradient is

[tex]\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}[/tex]

7. Set [tex]y=5[/tex] and solve for [tex]x[/tex]. The curve and line meet when

[tex]5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92[/tex]

Compute the derivative (for the curve) and evaluate it at these [tex]x[/tex] values.

[tex]\dfrac{dy}{dx} = 4x + 7[/tex]

[tex]\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}[/tex]

[tex]\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}[/tex]

8. Compute the derivative.

[tex]y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b[/tex]

The gradient is 8 when [tex]x=2[/tex], so

[tex]2a\cdot2 + b = 8 \implies 4a + b = 8[/tex]

and the gradient is -10 when [tex]x=-1[/tex], so

[tex]2a\cdot(-1) + b = -10 \implies -2a + b = -10[/tex]

Solve for [tex]a[/tex] and [tex]b[/tex]. Eliminating [tex]b[/tex], we have

[tex](4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}[/tex]

so that

[tex]4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}[/tex].

In a flower garden of rose and sunflower, 40% of flowers are rose. if there are 120 sunflower in the garden, total of how many flowers are there in the garden?​

Answers

Answer:

200

Step-by-step explanation:

If 40% are roses than 60% must be sunflowers (40% + 60% = 100%)

We are trying to find 60% of some number is 120.

Let n = the total number of flowers

.6n = 120  Divide both sides by .6

n = 200

You need to haul a load of patio bricks to a job site. Each brick weighs 4 pounds 14 ounces. Your truck can carry a 3/4 -ton load. How many bricks can your truck carry in a full load?

A.
300

B.
307

C.
362

D.
409

E.
483

Answers

Answer is B. 307.

Step by step.
We know there are 2000 pounds in a ton. So 3/4 of a ton can be found by multiplying 2000 x 3/4 = 1500. So your truck can carry 1500 pounds.

Now we need to divide 1500 by the weight of each brick to find out how many bricks you can carry.

1500 divided by 4 14/16 (14 over 16 ounces in a pound)
Change the fraction into a mixed fraction 4 x 16 + 14 = 78/16.

1500 divided by 78/16. To divide a fraction flip it and multiply.

1500 divided by 16/78 and multiply across you get 2400/78.

Now divide that to get your total bricks you can carry.

307.69. Since you can’t round up you have a total of 307 bricks.

Answer is B. 307 bricks.

Problem solved!

Triangle Congruency Theorems: Edge High School Geometry

Answer is below:

Answers

DE ≅ CE given that side AD and side BC are equal and angle ∠BCD and angle ∠ADC are equal. This can be obtained by using triangle congruency theorems.

Prove that  side DE and side CE is equal:

Triangle congruency theorems required in the question,

SAS triangle congruency theorem - SAS means Side-angle-side. If two sides and the included angle of a triangle is equal to two sides and the included angle of another triangle then the triangles are congruent.AAS triangle congruency theorem - AAS means Angle-angle-side. If two angles and one side of a triangle is equal to two angles and one side of another triangle then the triangles are congruent.

 

In the question it is given that,

⇒ Side DE and side CE are equal ⇒ AD ≅ BC

⇒ angle ∠BCD and angle ∠ADC are equal ⇒ ∠BCD ≅ ∠ADC

AD ≅ BC (given in the question)∠BCD ≅ ∠ADC (given in the question)DC ≅ DC (since DC is a common side; reflexive property)

Therefore we can say that,

ΔADC ≅ ΔBCD according to the SAS triangle congruency theorem

∠DAE ≅ ∠CBE (corresponding parts of congruent triangles are congruent (CPCTC))AD ≅ BC (given in the question)∠DEA ≅ ∠CEB  (since they are vertically opposite angles - vertical angles theorem)

ΔAED ≅ ΔBEC according to the AAS triangle congruency theorem

Thus DE ≅ CE since corresponding parts of congruent triangles are congruent (CPCTC).

Hence DE ≅ CE given that side AD and side BC are equal and angle ∠BCD and angle ∠ADC are equal.

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The Tortoise and the Hare is a fable about a race with the
moral, “Slow and steady wins the race.” The Tortoise and the
Hare decide to “race” across the United States from
Washington D.C. to Los Angeles.
a. Charles Darwin studied the tortoises when he was on
the Galapagos in 1835. He thought they moved relatively quickly. “One large one, I found by pacing, walked at the rate of 60 yards in 10 minutes” he wrote in Zoology Notes.
i How many inches per minute does the Galapagos tortoise walk?
ii. How long would it take a Galapagos tortoise to walk across the U.S.? Use the most
appropriate unit for time.
b. The Hare can run up to 30mph.
i. How many inches per minute does the Hare run?
ii. How long would it take the Hare to complete the race across the U.S.? Use the
most appropriate unit for time.
c. In the fable, the Hare, confident that he is so far head, relaxes and takes a nap. How long would the Hare need to nap for the Tortoise to pass him and win the race?

Answers

a.i) The number of inches per minute that the Galapagos tortoise walks are 216 inches (6 x 36).

a.ii) The time it would take a Galapagos tortoise to walk across the U.S. in hours is 2,933 hours (1,056,000/(6 x 60).

b.i) The Hare runs 31,680 inches per minute (1,900,800/60).

b.ii) The time it would take the Hare to complete the race across the U.S. in hours is 20 hours (600/30).

c.) For the Tortoise to pass the Hare, the Hare needs to sleep for 2,913 hours (2,933 - 20) or 121.4 days (2,933/24).

What are the equivalent values for calculating distance?

In this exercise, the following equivalent values are used for computing the distance and speed:

1 mile = 1,760 yards1 yard = 36 inches1 hour = 60 minutesDistance from Galapagos to the U.S. = 600 miles600 miles = 1,056,000 yards (600 x 1,760).

Data and Calculations:

Speed of Galapagos Tortoise per minute = 6 yards/ min (60 yards in 10 minutes)

Speed of Hare = 30mph

30 miles = 52,800 yards (30 x 1,760)

52,800 yards = 1,900,800 inches (52,800 x 36)

= 880 yards per minute (52,800/60)

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One fifth increase over 400

Answers

Answer:

480

Step-by-step explanation:

1/5 of 400 =   1/5 * 400 = 80

  increase 400 by 80   = 480

2. What is the length of the hypotenuse k?

Answers

Answer:

k ≈ 50.77

Step-by-step explanation:

using the cosine ratio in the right triangle

cos19° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{48}{k}[/tex] ( multiply both sides by k )

k × cos19° = 48 ( divide both sides by cos19° )

k = [tex]\frac{48}{cos19}[/tex] ≈ 50.77 ( to 2 dec. places )

Hi :)

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            We'll use sohcahtoa to solve this problem

[tex]\Large\boxed{\begin{tabular}{c|1} \sf{Sohcahtoa} ~&~~~~~Formula~~~~~~~ \\ \cline{1-2} \ \sf{Soh} & Opp~\div \text{hyp}\\\sf{Cah} & Adj \div \text{hyp}\\\sf{Toa} & Opp \div \text{adj} \end{tabular}}[/tex]

Looking at our triangle, we can clearly see that we have :

adj. side = 48 (adjacent to the angle)hyp. k (the one we need)

Set up the ratio

[tex]\longrightarrow\darkblue\sf{cos(19)=\dfrac{48}{k}}[/tex]

solve for k

[tex]\longrightarrow\darkblue\sf{k\cos(19)=48}[/tex] > multiply both sides by k to clear the fraction

[tex]\longrightarrow\darkblue\sf{k=\dfrac{48}{\cos(19)}}[/tex]  > divide both sides by cos (19)

[tex]\star\longrightarrow\darkblue\sf{k\approx50.77}\star[/tex]

[tex]\tt{Learn~More ; Work\ Harder}[/tex]

:)

on the graph, sketch f(x)=x+3 as well as g(x)=x

Answers

Answer:

below

Step-by-step explanation:

I need help please yall

use orders of operation and simplify ​

Answers

Answer: [tex]\large\boxed{12x^3-30x+3=0}[/tex]

Step-by-step explanation:

[tex]Solve:\\\ 6(2x^2-5)x=-3\\\\Use\ distributive\ property\ first\\\\(12x^2-30)x=-3\\\\Now\ multiply\ the\ terms\ in\ ()\ by\ 'x'\\\\12x^3-30x=-3\\\\\large\boxed{12x^3-30x+3=0}[/tex]

NO LINKS! Please help me with this problem

What is the standard form of the equation of the ellipse? ​

Answers

Answer:

[tex]\frac{x^2}{9}+\frac{y^2}{4}=1[/tex]

Step-by-step explanation:

So an ellipse can be expressed in two different but very similar forms:

Horizontal Major Axis:

   [tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]

Vertical Major Axis:
   [tex]\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1[/tex]

In both equations the length of the major axis is "2a" and the length of the minor axis is "2b"

In the equation you provided, the major axis is horizontal, and the minor axis is vertical.

So looking at the horizontal length, you can see that it's 6, and since this ellipse has a horizontal major axis, that means 2a=6, which means a=3

Looking at the vertical length, you can see that it's 4, and since the ellipse has a vertical minor axis, that means 2b=4, which means b=2

The last thing to note is that the center of an ellipse is (h, k) in the equation, and since here the center is (0, 0) it's (x-0)^2 and (y-0)^2 in the denominator which is just x^2 and y^2

So let's plug in the values into the equation:

[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]

(h, k) = (0, 0)

a = 3

b = 2

[tex]\frac{x^2}{3^2}+\frac{y^2}{2^2}=1\\\\\frac{x^2}{9}+\frac{y^2}{4}=1[/tex]

Here we go ~

The given figure is of a horizontal ellipse with :

Length of major axis = 2a = 6

so, a = 3

Length of minor axis = 2b = 4

hence, b = 2

And as it's shown in the figure, the ellipse has its centre at origin, so we can write it's equation as :

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{ {a}^{2} } + \cfrac{ {y}^{2} }{b {}^{2} } = 1[/tex]

[ now, plug in the values ]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{ {3}^{2} } + \cfrac{ {y}^{2} }{2 {}^{2} } = 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{ {9}^{} } + \cfrac{ {y}^{2} }{4 {}^{} } = 1[/tex]

That's all, ask me if you have any questions ~

A data set contains three points, and two of the residuals are -10 and 20.
What is the third residual?

Answers

If a data set contains three points, and two of the residuals are -10 and 20, the third residual is 10 (option B).

What is a residual?

A residual is the difference between the observed value and the estimated value of the quantity of interest.

The residual of a data points should normally sum up to zero (0). This means the following applies:

-10 + 20 + x = 0

x = 10

Therefore, if data set contains three points, and two of the residuals are -10 and 20, the third residual is 10.

Learn more about residual at: https://brainly.com/question/27864993

#SPJ1

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