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nataly862011 [7]

4 years ago

5

A mountain climber descends 3,852 feet over a period of 4 days. What was the average amount of her descent over that period of t

ime?

2 answers:

Crank4 years ago

7 0

Average = (total fee)/(total days)

Average= 3852/4 = 963 ft/day

Harrizon [31]4 years ago

5 0

To get the answer, divide 3852 feet by 4 days. That will give you an average of 963 feet per day as your answer.

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I have $100. I spent $40 on a prize and $45 on food. How do I figure out what % is purchase is from the original amount I had?

Paladinen [302]

You could say that you spent (purchased) 85% of the money that you had.

You spent a total of $85.

If you create the fraction 85/100, that is the same as saying you spent 85% of the money.

8 0

4 years ago

Brittney wants to buy a dress that costs $31.50, not including sales tax. The final cost of the dress, with sales tax included i

Fittoniya [83]

$1.89 all you have to do is subtract the final cost to the cost before taxes.

5 0

4 years ago

Expand and simplify: 2(x-4^2)+7

Mekhanik [1.2K]

Answer:
2x-25

Explanation:
2(x-4^2)+7
1) Find the exponent
=2(x-16)+7
2) Use the distributive property on the parentheses
=2x-32+7
3) Subtract 32 and 7
=2x-(32-7)
=2x-25

I hope this helps! Please comment if you have any questions.

3 0

3 years ago

Given: m∠APC = 27°, m DC = 27° Find: m AB

Wittaler [7]

Answer:

The unknown angle = 126°

Step-by-step explanation:

From angles in a triangle.

The total sum of angles in a triangle = 180°

Let the unknown angle be x

27 + 27 + x = 180

X = 180-27-27

X= 180-54

X = 126°

6 0

3 years ago

Read 2 more answers

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit theorem

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago