Evaluate the expression. StartFraction 9 factorial Over 3 factorial EndFraction

Jack and Jill had 3 pails of water. Each trip up the hill, Jack and Jill obtained 3 more buckets of water. Which of the graphs b

lana66690 [7]

This is the graphed answer for this problem

7 0

2 years ago

the perimeter of an isosceles trapezoid is 62 cm. if three sides are equal in length and the fourth side is 10 cm longer, find t

Alja [10]

Perimeter = sum of all the sides
let x = one side
longer side = x + 10
Perimeter = x + x + x + x + 10
62 = 4x + 10
4x = 52
x = 13
Longer side = 23

Area = (10+23)x2/10
= 6.6

5 0

2 years ago

Read 2 more answers

A solid lies between two parallel planes 22 feet apart and has a volume of 4343 cubic feet. What is the average area of cross-se

ella [17]

Answer:

Step-by-step explanation:

So, formula to be used is:

Average area of cross section is = volume/distance between plates.

Volume = 4343/22 = 197.4 square feet.

8 0

2 years ago

Ecologists sometimes find rather strange relationships in our environment. For example, do beavers benefit beetles? Researchers

Harman [31]

Answer:

The appropriate conclusion that answers the question stated in step one is:

The weak correlation value does not support the idea that beavers benefit beetles.

The largest number of beetle larvae was 55 and the largest number of stumps was 5.

Step-by-step explanation:

1) Select the appropriate practical question.

The appropriate practical question is:

"How does the number of beaver-caused stumps influences the number of beetle larvae clusters?"

2) Make a scatter plot.

The scatter plot is attached below.

From the scatter plot it can be seen that there exists a negative relationship between beaver stumps and beetle larvae. That is as the number of stumps increases the number of beetle larvae decreases.

The maximum number of beaver stumps found is 5.

3)

The best numerical summary for this problem are,

Correlation and regression line

The least square regression line for predicting the number of beetle larvae from the number of stumps is:

y = 31.594 - 2.8365 x.

Use the Excel function "=CORREL(array1,array2)" to compute the correlation.

The correlation value is, r = -0.219.

Both the correlation coefficient and regression line explains a negative relationship between the number of beetle larvae and the number of stumps.

Values amid 0 and 0.3 (0 and -0.3) implies a weak positive (negative) linear relationship amid the variables.

Thus, a correlation coefficient, r = -0.219, indicates a weak negative correlation between two variables.

Thus, the appropriate conclusion that answers the question stated in step one is:

The weak correlation value does not support the idea that beavers benefit beetles.

The largest number of beetle larvae was 55 and the largest number of stumps was 5.

3 0

2 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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