The scores of students on a standardized test are normally distributed with a mean of 300 and a standard deviation of 40. (a) what proportion of scores lie between 220 and 380 points?

Answer:

okay it's name is Muhammad Deco alfansia ( ◜‿◝ )

Answer:

Slope, sometimes referred to as gradient in mathematics, is a number that measures the steepness and direction of a line, or a section of a line connecting.

Step-by-step explanation: Hope this helps you!

The value of a₃₁ of the arithmetic sequence exists 77.4.

Given: a₅ = 12.4 and a₉ = : 22.4

For the arithmetic sequence a₁, a₂, a₃, ..., the n-th term exists

where d = common difference

a₅ = 12.4,

a₁ + 4d = 12.4 .........(1)

Because a₉ = 22.4,

a₁ + 8d = 22.4 .........(2)

Subtract (1) from (2), we get

a₁ + 8d - (a₁ + 4d) = 22.4 - 12.4

4d = 10

Dividing throughout by 4, we get

d = 2.5

From (1), we get

a₁ = 12.4 - 4 [tex]*[/tex] 2.5 = 2.4

a₃₁ = 2.4 + 30 [tex]*[/tex] 2.5 = 77.4

Therefore, the correct answer is a₃₁ = 77.4

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Nag basa nito walang jõwa

Answer:

x = 55°

Step-by-step explanation:

x , 35° , 90° lie on a straight line and sum to 180° , that is

x + 35° + 90° = 180°

x + 125° = 180° ( subtract 125° from both sides )

x = 55°

Answer:

[tex]\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x=-\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

[tex]\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}[/tex]

[tex]\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}[/tex]

Using Integration by parts:

[tex]\textsf{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}[/tex]

[tex]\textsf{Let }\dfrac{\text{d}v}{\text{d}x}=x^{-8} \implies v=-\dfrac{1}{7}x^{-8+1}=-\dfrac{1}{7}x^{-7}=-\dfrac{1}{7x^7}[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x & = -\dfrac{1}{7x^7}\ln x- \int -\dfrac{1}{7x^7} \cdot \dfrac{1}{x}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7x^8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7}x^{-8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}-\dfrac{1}{7 \cdot 7}x^{-8+1}+\text{C}\\\\& = -\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}\end{aligned}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Differentiating $\ln x$}\\\\If $y=\ln x$, then $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1}{x}$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]

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Substitute [tex]y=\ln(x) \iff e^y = x[/tex] and [tex]dy=\frac{dx}x[/tex] to get

[tex]\displaystyle \int \frac{\ln(x)}{x^8} \, dx = \int ye^{-7y} \, dy[/tex]

Integrate by parts with

[tex]u = y \implies du = dy[/tex]

[tex]dv = e^{-7y} \, dy \implies v = -\dfrac17 e^{-7y}[/tex]

Then

[tex]\displaystyle \int u\,dv = uv - \int v\,du \\\\ \implies \int y e^{-7y} \, dy = -\dfrac17 ye^{-7y} + \frac17 \int e^{-7y} \, dy + C \\\\ ~~~~~~~~~~~~ = -\frac17 ye^{-7y} - \frac1{49} e^{-7y} + C \\\\ ~~~~~~~~~~~~ = \boxed{-\frac{\ln(x)}{7x^7} - \frac1{49x^7} + C}[/tex]

Answer:

7.07

Step-by-step explanation:

Answer:

7.07

Step-by-step explanation:

Hello!

Let's find two perfect square numbers that are directly before and after 50.

Since the square of 7 is the closest, we can use that as our whole number.

49 is 1 away from 50, and 64 is 14 away. Rewriting it as a fraction and we get  [tex]\frac{1}{14}[/tex].  The decimal is 0.07142857142.

Now, put 7 and 0.07142857142 together and we get 7.07142857142. Rounding that, we get 7.07.

The real value of Root 50 is 7.071067811865475, so the decimals were really close.

Answer:

Discount =40\%

Selling price of headphone =\$36

Solution --

we know that,

MP=SP * 100/(100-D\%)

putting values now we get,

MP = 36 * 100 / 60=60$

so, initial price of Headphone was =$60

The examples of exponential growth include bacteria population growth and compound interest and a real life example of exponential decay is radioactive decay.

Exponential growth is the pattern of data that shows sharper increases over time. Savings accounts with a compounding interest rate can show exponential growth.

There are many real-life examples of exponential decay. An example, is thatsuppose that the population of a city was 100,000 in 1980. Then every year after that, the population has decreased by 3% as a result of heavy pollution. This is an example of exponential decay.

One of the best examples of exponential growth is the observed in bacteria. It takes bacteria roughly an hour to be able to reproduce through prokaryotic fission. In this case, if we placed 100 bacteria in an environment and recorded the population size each hour, we would observe an exponential growth.

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Answer:309

Step-by-step explanation:

Step-by-step explanation:

Total Angle in a triangle is 180°

so D = 140 + 25 = 165°

D = 190° - 165° = 15°

Your answer is 15°.

Answer:

angle d = [tex]\boxed{15}^ {\circ}[/tex]

Step-by-step explanation:

The angles in a triangle add up to 180°.

∴ ∠d + 25° + 140° = 180°

⇒ ∠d + 165° = 180°

⇒ ∠d = 180° - 165°

⇒ ∠d = 15°

The numerical expression, 2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3 = 67/24 on simplification using the BODMAS rule.

In the question, we are asked to simplify the numerical expression:

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3.

To simplify the expression, we will follow the BODMAS rule, where B means Brackets, O means Of, D means Divide, M means Multiplication, A means Addition, and S means Subtraction.

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3

= 2/3 ÷ 16 + (3/4 + 1/6) ÷ 1/3 {Solving 2⁴ = 16, before proceeding BODMAS}.

= 2/3 ÷ 16 + ((9+2)/12) ÷ 1/3 {Solving Brackets by taking LCM}

= 2/3 ÷ 16 + 11/12 ÷ 1/3 {Simplifying}

= 2/3 * 1/16 + 11/12 * 3/1 {Solving divisions by taking reciprocals}

= 1/24 + 11/4 {Multiplying}

= (1 + 66)/24 {Adding using LCM}

= 67/24 {Simplifying}.

Thus, the numerical expression, 2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3 = 67/24 on simplification using the BODMAS rule.

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The provided question is incomplete. The complete question is:

"Type the correct answer in the box. Use numerals instead of words. For this item, a non-integer answer should be entered as a fraction using / as the fraction bar.

Simplify the numerical expression.

2/3 ÷ 2⁴ + (3/4 + 1/6) ÷ 1/3

The expression has a value equal to."

Based on the given parameters of a = 1.7 cubic feet and concrete for first 4 steps as 17 cubic feet, the concrete is required for the first 12 steps is 132.6 cubic feet

Sn = n/2 {2a + (n - 1) d}

17 = 4/2{2×1.7 + (4 - 1)d}

17 = 2{3.4 + (3)d}

17 = 2(3.4 + 3d)

17 = 6.8 + 6d

17 - 6.8 = 6d

10.2 = 6d

d = 10.2/6

Concrete required for first 12 steps;

Sn = n/2 {2a + (n - 1) d}

= 12/2{2×1.7 + (12-1)1.7}

= 6{3.4 + (11) 1.7}

= 6(3.4 + 18.7)

= 6(22.1)

= 132.6

Concrete required for first 12 steps = 132.6 cubic feet

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Answer:

132.6 cubic feet

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

It's B

Answer:

Option B

Step-by-step explanation:

The equation is:

[tex]y=10-2x[/tex]

when x=2

[tex]y=10-2(2)010-4=6[/tex]

When x = 3

[tex]y=10-2(3)=10-6=4[/tex]

When x=4

[tex]y=10-2(4)=10-8=2[/tex]

Hope this helps

The distance between points S' and S is 2.5 units.

If a line is drawn parallel to any one side of a triangle so that it intersects the other two sides in two distinct points, then the other two sides of the triangle are divided in the same ratio.

Given that,

line segment ST is dilated to create line segment S'T' using the dilation rule DQ.

Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5 units, SS' = x units.

We need to find the value of x, the distance between points S' and S.

Since the line ST is dilated to S'T' with center of dilation Q, so the triangles STQ and S'T'Q must be similar.

We know that the corresponding sides of two similar triangles are proportional.

So, from ΔSTQ and ΔS'T'Q, we get

[tex]\frac{SQ}{S'Q} =\frac{TQ}{T'Q}[/tex]

[tex]\frac{SQ}{SQ+S'S} =\frac{TQ}{TQ+T'Q}[/tex]

[tex]\frac{2}{2+x} =\frac{1.2}{1.2+1.5}[/tex]

[tex]\frac{2}{2+x} =\frac{1.2}{2.7}[/tex]

[tex]\frac{2}{2+x} =\frac{12}{27}[/tex]

[tex]\frac{1}{2+x} =\frac{6}{27}[/tex]

12+6x = 27

6x = 15

x = 2.5

Hence, Thus, the required value of x is 2.5 units.

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The original amount of ida's rent is $56.

First consider the unknown original amount as 'x'. To find the increase or decrease, multiply the rate by the original amount 'x'. To find the final amount, add or subtract the increase or decrease to the original amount 'x' and equate this to given final amount. Solve the equation and find the original amount 'x'.

Let the original amount be x. This amount increase by 5% after increasing the amount is $59.

[tex]x+x(\frac{5}{100})=59\\\\[/tex]

[tex]x+x(\frac{1}{20})=59\\\\[/tex]

[tex]\frac{21x}{20}[/tex]  = 59

x = $56.19

Hence, The original amount be x =$56.19.

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Answer:

600 and 200

Step-by-step explanation:

kofi  : kweku     is   3 :1

   so Kofi gets   3 out of ( 3 +1)  = 3/4 of 800  = 3/4 * 800 = 600

         kweku get s the rest   800- 600 = 200  

The first quartile of the dataset {1, 2, 3, 4, 5, 6} is 2.

Given data set: {1,2,3,4,5,6}

Median of the given data set =3.5

The data set can be split into 2 parts as, {1,2,3} and {4,5,6}. The first quartile is calculated by simply finding the median of the first part of the data set.

Thus, the first quartile is 2.

The quarter divides the distribution into four groups and calculates the range of values above and below the mean.

A quartile separates the dataset into four categories by dividing the data into three points: the lowest, median, and upper quartiles.

The interquartile range, a measurement of variation around the median, is calculated using quartiles.

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Answer:

7

Step-by-step explanation:

You take the corresponding side that you know which is 10.5 and you multiply that by your scale factor of 2/3.

Another name for 10.5 is 10 [tex]\frac{1}{2}[/tex]  and that can be changed to [tex]\frac{21}{2}[/tex]

([tex]\frac{21}{2}[/tex])([tex]\frac{2}{3}[/tex])  The two's cancel out and we are left with [tex]\frac{21}{3}[/tex]  Which is the same as 7.

In quadrant IV, [tex]\cos(A)[/tex] is positive. So

[tex]\sin^2(A) + \cos^2(A) = 1 \implies \cos(A) = \sqrt{1-\sin^2(A)} \approx 0.6258[/tex]

Then by the definition of tangent,

[tex]\tan(A) = \dfrac{\sin(A)}{\cos(A)} \approx \dfrac{-0.78}{0.6258} \approx \boxed{-1.2465}[/tex]

Answer:

540

Step-by-step explanation:

540 =18% of 97.2

Done please

it's quotient it's a synthax error..

[tex]g(x)=5cos((\pi )/(2)x-(3\pi )/(2))-2[/tex]

generate by:  Amplitude:5    Period:4

                      Phase shift:(3 to the right)    Vertical shift:-2

x=3,g(x)= 3

x=4,g(x)= -2

x=5,g(x)= -5

x=6,g(x)= -2

x=7,g(x)= 3

the graph is like cos(x)

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The length of the radius of a circle exists 41 units.

Given: The center exists at -7+2i and a point in the circle at 33+11i.

The radius of the circle exists given by the following formula;

The radius of the circle [tex]$=\sqrt{x^{2}+y^{2}}$[/tex]

The center exists at -7 + 2i and a point in the circle at 33 + 11i.

[tex]$&x(33-(-7)), y(2 \mathrm{i}-11 \mathrm{i}) \\[/tex]

simplifying the equation, we get

[tex]$&\mathrm{x}(33+7), \mathrm{y}(-9 \mathrm{i}) \\[/tex]

[tex]$&\mathrm{x}(40), \mathrm{y}(-9(-1)) \\[/tex]

[tex]$&\mathrm{x}(33+7), \mathrm{y}(9)[/tex]

The center of the circle exists at [tex]$&\mathrm{x}(33+7), \mathrm{y}(9)[/tex].

The length of the radius of a circle exists,

Radius [tex]$}=\sqrt{x^{2}+y^{2}} \\[/tex]

substituting the values of x and y, we get

Radius[tex]$}=\sqrt{40^{2}+9^{2}} \\[/tex]

Radius [tex]$}=\sqrt{1600+81} \\[/tex]

Radius [tex]$=\sqrt{1681} \\[/tex]

Radius = 41 unit

Therefore, the length of the radius of a circle exists 41 unit.

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Answer:

A = (0, b)

B = (a, b)

Step-by-step explanation:

A is straight up from the origin and is the same height as b so it's (0, b)

B is straight up from C so the x is the same so it's (a, b)

Answer:

QP = | a - d |

Step-by-step explanation:

since the y- coordinates of P and Q are equal , both b

then PQ is the absolute value of the difference of the x- coordinates, that is

QP = | a - d | = | d - a |

Answer: the area of the triangle is 5.

Step-by-step explanation:

[tex]A(0;0) \ \ \ \ B(1;3) \ \ \ \ C(4;2) \ \ \ \ S_{ABC}=?\\Use \ the\ formula:\\\displaystyle\\\boxed{S=\frac{1}{2}*|[(x_A-x_C)*(y_B-y_C)-(x_B-x_C)*(y_A-y_c)] |}\\x_A=0\ \ \ \ x_B=1\ \ \ \ x_C=4\ \ \ \ y_A=0\ \ \ \ \ y_B=3\ \ \ \ \ y_C=2.\\S=\frac{1}{2}*|[(0-4)*(3-2)-(1-4)*(0-2)]|\\S=\frac{1}{2}*| [(-4)*1-(-3)*(-2)]|=\\ S=\frac{1}{2}*| (-4-6)|\\S=\frac{1}{2}*|(-10)|\\S= \frac{1}{2} *10\\S=5.[/tex]

The approximate year at which two revenues were equal is; 7.5 years

From the graph attached, we see that in the year 10, the revenue for printed ads was 2 million dollars & 3 million dollars for printed ads and online ads revenue respectively.

Thus, printed ad. Revenue line equation is;

(y - 2) = (3 - 2)(x - 10)/(0 - 10)

y - 2 = (x - 10)/-10

x - 10 = -10(y - 2)

x - 10 = -10y + 20

x - 10 = -10y + 20

x  + 10y = 30   -----(1)

At x = 12 years from the graph, we have;

12 + 10y = 30

10y = 18

y = 1.8

Thus, online ad. Revenue line equation is;

(y - 0) = ((3 - 0)/(10 - 0))(x - 0)

y = 3x/10

10y = 3x

10y - 3x = 0 -----(2)

At x = 12, we have;

10y = 3*12

10y = 36

y = 3.6

In year '12' the online ad revenue got doubled as that of printed ad revenue and afterward more than doubled.

B) The approximate year at which two revenues were equal is gotten by solving equation 1 and 2 simultaneously to get;

x = 7.5

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[tex]7x^{2} - 9x^{2} +8x-5x + 10-7[/tex] is the given expression

Like terms are terms whose variables (and their exponents such as the 2 in [tex]x^{2}[/tex]) are the same. Like terms can easily be added and substracted. Such questions just need like terms to be bought together and simplified

Infinite number of equations or expressions can be written for the given question. So writing one possibility for the given question,

= [tex]7x^{2} - 9x^{2} +8x-5x + 10-7[/tex]

which sums up to

= [tex]-2x^{2} +3x+3[/tex]

Thus [tex]7x^{2} - 9x^{2} +8x-5x + 10-7[/tex] is the given expression

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Answer:

  (f +g)(2) = -7

Step-by-step explanation:

Functions are added by adding their values.

  (f +g)(2) = f(2) +g(2)

  = (2 -8) +(4(2) -9) . . . . . evaluate f and g with z=2

  = -6 +(-1)

  (f +g)(2) = -7