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Irina-Kira [14]

3 years ago

15

Elise goes to an exercise class every Wednesday night. 45% of the students in the class are men. If there are 20 students in the

class, how many men are in Elise's exercise class?

1 answer:

a_sh-v [17]3 years ago

3 0

There are 9 men in the class

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Surface area... help me

alexandr402 [8]

Answer: Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces. We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.

Step-by-step explanation:

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2 years ago

Paige has 3 rain coats, a pink one, a red one, and a yellow one. She has blue rain boots, green rain boots, and 1 yellow umbrell

Harlamova29_29 [7]

The combination is 4/12

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3 years ago

Use the following table for questions 11-13.A baseball manager believes a linear relationship exists between the number of Home

slamgirl [31]

The regression equation of Y on X is given by the following formula:

$Y-\bar{Y}=b_{yx}(X-\bar{X})$

Where byx is given by the formula:

$b_{yx}=\frac{N\sum^{}_{}XY-\sum^{}_{}X\sum^{}_{}Y}{N\sum^{}_{}X^2-(\sum^{}_{}X)^2}$

Where N is the number of values (N=8). We need to find the sum of X values, the sum of Y values, the average of X, the average of Y, the sum of X*Y and the sum of X^2.

The table of values is:

The values we need to know are on the following table:

By replacing the known values in the formula we obtain:

$\begin{gathered} b_{yx}=\frac{8\cdot26125-167\cdot995}{8\cdot4649-(167)^2} \\ b_{yx}=\frac{209000-166165}{37192-27889} \\ b_{yx}=\frac{42835}{9303} \\ b_{yx}=4.6 \end{gathered}$

Now, the average of X and Y is the sum divided by N, then:

$\begin{gathered} \bar{X}=\frac{167}{8}=20.87 \\ \bar{Y}=\frac{995}{8}=124.37 \end{gathered}$

Replace these values in the formula and find the regression equation as follows:

$\begin{gathered} Y-124.37=4.6(X-20.87) \\ Y-124.37=4.6X-4.6\cdot20.87 \\ Y=4.6X-96.11+124.37 \\ Y=4.6X+28.26 \end{gathered}$

The answer is a) y=4.6x+28.26

7 0

1 year ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when $X = 59.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{59.5 - 56}{1.12}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

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navik [9.2K]

Have a nice day!!!!

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3 years ago