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4 years ago

Use the circle graph to determine how many hours per day Becky spends on each activity.

1 answer:

3 0

404 error: graph not found

anyway, the graph was not included. since the sleep time was included, I will assume that the circle graph is worth 24 hours

all we need to do is to convert the percentages to fractions and multiply that by 24 to find out how many hours per activity

percent means parts out of 100 so x%=x/100

so we have
School
Eating
Sleep
Homework
Free Time

School=25%
25%=25/100=1/4
1/4 times 24=6
School: 6 hours

Eating=10%
10%=10/100=1/10
1/10 times 24=2.4
Eating: 2.4 hours

Sleep=40%
40%=40/100=4/10=2/5
2/5 times 24=48/5=9.6
Sleep: 9.6 hours

Homework=10%
10%=10/100=1/10
1/10 times 24=2.4
Homework: 2.4 hours

Free Time=15%
15%=15/100=3/30
3/20 times 24=72/20=36/10=3.6
Free Time: 3.6 Hours

Answers:
School: 6 hours
Eating: 2.4 hours
Sleep: 9.6 hours
Homework: 2.4 hours
Free Time: 3.6 Hours

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3 years ago

A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye

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Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{x}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{x}=\sqrt{n}\sigma$

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

$P (\sum X \leq 100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}$

$=P(Z$

*Use a z-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

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3 years ago