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Aleonysh [2.5K]

3 years ago

14

A store sells four-packs of permanent markers in assorted colors for $3.19 per pack. All spiral-bound notebooks, whether wide-ru

led or college-ruled, sell for $1.20 each. Cleo buys x packs of markers, y wide-ruled notebooks, and z college-ruled notebooks. The store charges a 6% sales tax on Cleo’s total bill. Which expression represents the amount Cleo must pay after taxes?

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer: (x3.19+y1.20 + z1.20) x 1.06

Step-by-step explanation:

Hi, to answer this question we have to write an expression.

Four-packs of permanent markers in assorted colors for $3.19 per pack.

3.19 x (x number of packs purchased)

spiral-bound notebooks, whether wide-ruled or college-ruled, sell for $1.20 each

1.20 y + 1.20 z (y wide-ruled notebooks, and z college-ruled notebooks)

The store charges a 6% sales tax on Cleo’s total bill, so we have to multiply the expression by 1+ 0.06 ( 1+ tax percentage in decimal form)

(x3.19+y1.20 + z1.20) x 1.06

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<u><em>Answer: ⇒⇒⇒⇒⇒⇒⇒ =54</em></u>

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Step-by-step explanation:

First you had to convert element to fraction.

$\frac{2}{3}*\frac{81}{1}$

Then you had to cross out by the common factor of 3.

$\frac{2}{1}=\frac{27}{1}$

Multiply by the fraction.

$\frac{2*27}{1*1}$

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a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be 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 $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3$

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

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

X = 7.1

$Z = \frac{7.1 - 6.3}{0.3}$

$Z = 2.67$

$Z = 2.67$ has a pvalue of 0.9962

X = 5.5

$Z = \frac{5.5 - 6.3}{0.3}$

$Z = -2.67$

$Z = -2.67$ has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

$Z = \frac{6.7 - 6.3}{0.3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

X = 5.9

$Z = \frac{5.9 - 6.3}{0.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

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