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3 years ago

I need someone to solve for x and explain their steps, please!

Mathematics

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

ok that is the definition of work that i canot complete

Step-by-step explanation:

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The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0

3 years ago

Can y’all help with this I don’t understand it

Westkost [7]

The answer is $990. Because 0.36 times 2750=990

4 0

3 years ago

Read 2 more answers

jenny is painting a toy chest that she made for her little brother. if the box is 4 feet long, 3 feet wide, and 2 feel tall, wha

horsena [70]

You would have to multiply all the numbers. 4 X 3, 4 X 2, 3 X 2, the answer I got was 36.

7 0

3 years ago

Choose a point uniformly at random in a unit square (square of side length 1). Let X be the distance from the point chosen to th

Montano1993 [528]

Answer:

The solution is attached below:

3 0

4 years ago

2 points) Test each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the ser

alexgriva [62]

Answer:

Required result : (a) Divergent. (b) Convergent. (c) Divergent, (d) Not exists. (e) Not exists.

Step-by-step explanation:

By the definition of integral test the series for a integer n and a continuous function f(x) which is monotonic decreasing in $[n,\infty)$ then the infinite series $\sum_{n}^{\infty}$ converges or diverges if and only if the improper integral $\int_{n}^{\infty}$ converges or diverges.

Given,

(a) $\sum_{n=1}^{\infty}9^n(\ln(9^n))^n$

$=5(\ln(9))^5\sum_{n=1}^{\infty}9^nn^5\hfill (1)$

Then,

$\int_{1}^{\infty}9^n(\ln(9^n))^5$

$=5(\ln 9)^5\int_{1}^{\infty}9^nn^5dn\hfill (2)$

Let,

$I=\int 9^nn^9$

By using integral calculator we get,

$I=\dfrac{\left(2\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(2\ln\left(3\right)n-5\right)+10\right)-15\right)+15\right)-15\right)\mathrm{e}^{2\ln\left(3\right)n}}{8\ln^6\left(3\right)}$