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3 years ago

jane had a fourth idea for her allowance:Give jane $0.50 in january then multiply her allowance by 4 each month

Mathematics

1 answer:

Alika [10]3 years ago

6 0

Answer:First, identify what the question is asking for.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse . How much was the blouse?

Next, identify the numbers.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse . How much was the blouse?

Next, identify the key words. These include add, subtract, remove, spend, earn, less, more, times, twice, half, etc.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse . How much was the blouse?

Finally, convert everything into an equation.

⋅

blouse

−

Now, solve the equation.

⋅

The blouse was 28 dollars.

Step-by-step explanation:

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0.625 is the answer to this

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each hour the number of bacteria in dr. nall's petri dish tripled what percent is the population compared to the population the

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3 years ago

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.

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