3 five-dollar bills, 3 one-dollar bills, 3 quarters, 1 dime, and 3 nickels. How much money does he have?

Two trains leave stations 374 miles apart at the same time and travel toward each other. One train travels at 95 miles per hour

sveta [45]

2 hrs and 12 min or 2.2 hrs

5 0

3 years ago

A student committee is to consist of 2 freshmen, 5 sophomores, 4 juniors, and 3 seniors. If 6 freshmen, 13 sophomores, 8 juniors

mezya [45]

<h3>Answer: 491,891,400</h3>

Delete the commas if necessary.

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Explanation:

There are 6 freshmen total and we want to pick 2 of them, where order doesn't matter. The reason it doesn't matter is because each seat on the committee is the same. No member outranks any other. If the positions were labeled "president", "vice president", "secretary", etc, then the order would matter.

Plug n = 6 and r = 2 into the nCr combination formula below

$n C r = \frac{n!}{r!(n-r)!}\\\\6 C 2 = \frac{6!}{2!*(6-2)!}\\\\6 C 2 = \frac{6!}{2!*4!}\\\\6 C 2 = \frac{6*5*4!}{2!*4!}\\\\ 6 C 2 = \frac{6*5}{2!}\\\\ 6 C 2 = \frac{6*5}{2*1}\\\\ 6 C 2 = \frac{30}{2}\\\\ 6 C 2 = 15\\\\$

This tells us there are 15 ways to pick the 2 freshmen from a pool of 6 total.

Repeat those steps for the other grade levels.

n = 13 sophomores, r = 5 selections leads to nCr = 13C5 = 1287. This is the number of ways to pick the sophomores.

You would follow the same type of steps shown above to get 1287. Let me know if you need to see these steps.

Similarly, 8C4 = 70 is the number of ways to pick the juniors.

Lastly, 14C3 = 364 is the number of ways to pick the seniors.

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To recap, we have...

15 ways to pick the freshmen
1287 ways to pick the sophomores
70 ways to pick the juniors
364 ways to pick the seniors

Multiply out those values to get to the final answer.

15*1287*70*364 = 491,891,400

This massive number is a little under 492 million.

7 0

2 years ago

A holiday decoration involves blue and white lights the ratio of blue lights to white lights is three to two the box the decorat

Westkost [7]

Answer:

15

Step-by-step explanation:

4 0

3 years ago

Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa

Debora [2.8K]

Answer:

a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

b) 0.4129 = 41.29% probability that the mean return will be less than 8%

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 estabilishes 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.

Mean 8.7% and standard deviation 20.2%.

This means that $\mu = 8.7, \sigma = 20.2$

40 years:

This means that $n = 40, s = \frac{20.2}{\sqrt{40}}$

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?

This is 1 subtracted by the pvalue of Z when X = 13. So

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

By the Central Limit Theorem

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