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ivanzaharov [21]

2 years ago

15

Jalen bought 24 juice packs for $7.20. Tia bought 6 juice packs for $2.70. Which statement describes the difference in the unit

prices of Jalen's and Tia's juice packs?

2 answers:

-BARSIC- [3]2 years ago

6 0

Answer:

$0.15 .

Step-by-step explanation:

Jalen bought 24 juice packs for $7.20

Cost per pack of juice = $\frac{7.20}{24}$

= $0.30

Tia bought 6 juice packs for $2.70

Cost per pack of juice = $\frac{2.70}{6}$

= $0.45

The difference in unit prices of Jalen's and Tia's juice packs

= 0.45 - 0.30 = $0.15

The difference in unit prices of Jalen's and Tia's juice packs is $0.15 .

Dafna1 [17]2 years ago

5 0

7.20/24 would give us 0.30/1 so Jalen's unit rate is $0.30 per juice pack.
Then we take our 2.70/6 which would give us 0.45/1 which means the unit rate is $0.45 per juice pack.What you're doing is trying to find the difference, so you'd subtract .30 from .45 which gives you .15, the difference is $0.15 per unit.

0.15 is your answer

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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil

MArishka [77]

Answer:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,0.3)$

Where $\mu=12$ and $\sigma=0.3$

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}$

If we solve for the total T we got:

$T= n \bar X$

For this case then the expected value and variance are given by:

$E(T) = n E(\bar X) =n \mu$

$Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2$

And the deviation is just:

$Sd(T) = \sqrt{n} \sigma$

So then the distribution for the total would be also normal and given by:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

And we want this probability:

$P(T\leq 72)$

And we can use the z score formula given by:

$z = \frac{x-\mu}{\sigma}$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

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