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Allisa [31]
3 years ago
12

At bjs you can buy 6 cans for $4.99 or 10 cans for 7.99. Which pack is a better buy?

Mathematics
2 answers:
WINSTONCH [101]3 years ago
7 0

For this, you have to calculate the Unit value,

Unit value of 1rst one,  4.99/6 = 0.83

Unit value of 2nd one, 7.99/10 = 0.79

As, 2nd one is cheaper, so it is better :))

Olegator [25]3 years ago
3 0
<h3>Answer with Step-by-step explanation:</h3>

You can buy:

6 cans for $4.99

i.e. 1 can for $4.99/6

i.e. 1 can for $0.83

or 10 cans for $7.99

i.e. 1 can for $7.99/10

i.e. 1 can for $0.799

Since, cost of 1 can on buying 10 cans for $7.99 is less than the case on buying 6 cans for $4.99

<h3>Hence, Pack which is better is:</h3><h3>10 cans for $7.99</h3>
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Step-by-step explanation:

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3 years ago
A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a m
andrezito [222]

Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand 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 = 2000, \sigma = 18

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So n = 100, s = \frac{18}{\sqrt{100}} = 1.8

This probability is the pvalue of Z when X = 1995. So

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

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Z = -2.78 has a pvalue of 0.0027

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This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when Z = 1.28

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X - 2000 = 1.8*1.28

X = 2002.3

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

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