All categories

3 years ago

Need help with this!!

Mathematics

1 answer:

Harrizon [31]3 years ago

7 0

Answer:

Step-by-step explanation:

i have never done that:(

You might be interested in

A film distribution manager calculates that 4% of the films released are flops. If the manager is correct, what is the probabili

Anit [1.1K]

Answer:

9.34%

Step-by-step explanation:

p = 4%, or 0.04

n = Sample size = 667

u = Expected value = n * p = 667 * 0.04 = 26.68

SD = Standard deviation = $\sqrt{np(1-p)} =\sqrt{667*0.04*(1-0.04)}$ = 5.06

Now, the question is if the manager is correct, what is the probability that the proportion of flops in a sample of 667 released films would be greater than 5%?

This statement implies that the p-vlaue of Z when X = 5% * 667 = 33.35

Since,

Z = (X - u) / SD

We have;

Z = (33.35 - 26.68) / 5.06

Z = 1.32

From the Z-table, the p-value of 1.32 is 0.9066

1 - 0.9066 = 0.0934, or 9.34%

Therefore, the probability that the proportion of flops in a sample of 667 released films would be greater than 5% is 9.34%.

6 0

3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

4 years ago

Read 2 more answers

Which of the following expressions have the greatest value

rusak2 [61]

Answer:

the Answer is D

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

I need help with these .

ollegr [7]

1. 102°
2. 56°
3. 104°
4. 138°
5. 95°
6. 133°

6 0

3 years ago

Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = x^3 – 3x^2 + 2x

castortr0y [4]

Answer: The correct option is second, i.e. ,"2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".

Explanation:

The given equation is,

$y=x^3-3x^2+2x$

It can be written as,

$f(x)=x^3-3x^2+2x$

Find the zeros of the equation. Equation the function equal to 0.

$0=x^3-3x^2+2x$

$x(x^2-3x+2)=0$

$x(x^2-2x-x+2)=0$

$x(x-2)(x-1)=0$

So, the three zeros are 0, 1 and 2.

The graph of the equation is shown below.

From the given graph it is noticed that the enclosed by the curve and x- axis is lies between 0 to 2, but the area from 0 to 1 lies above the x-axis and area from 1 to 2 lies below the x-axis. So the function will be negative from 1 to 2.

The area enclosed by curve and x-axis is,

$A=\int_{0}^{1}f(x)dx+\int_{1}^{2}[-f(x)]dx$

$A=\int_{0}^{1}f(x)dx-\int_{1}^{2}f(x)dx$

From the graph it is noticed that the area from 0 to 1 is symmetric or same as area from 1 to 2. So the total area is the twice of area from 0 to 1.

$A=2\int_{0}^{1}f(x)dx$

$A=2\int_{0}^{1}[x^3-3x^2+2x]dx$

Therefore, The correct option is "2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".

3 0

3 years ago