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3 years ago

14

Cases are normally sold full. There were three partially full cases that could be combined to make full cases. The cases were 11

/16, 5/16 and 7/16 full.

1 answer:

dsp733 years ago

5 0

11/16 + 5/16 + 7/16 = 23/16

23/16 = 1 7/16

There is enough for one full case and 7/16 case leftover.

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25 pnts ‼️‼️theres a picture for these. One would help Please need help due in two days forever ‼️‼️

ad-work [718]

There are a lot of math questions here. I'll answer the first 5 questions to get you started.

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Problem 19

This is a permutation because order matters and we want to find out the number of ways to arrange the people in the first, second, third and fourth place slots.

You would use the nPr formula.

=============================

Problem 20

We use the counting principle here. This is because we can create a tree diagram. To find the number of branches in this diagram, you would multiply out all the values.

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Problem 21

This is similar to problem 19; however, order doesnt matter in this case. All that matters is the grouping of books. For example, the group {A,B,C} is the same as {B,C,A} because both groups have the same books in them and the the groups have the same number of books. So in short, we use a combination here instead of a permutation.

You would use the nCr formula.

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Problem 22

Similar to problem 21. Order does not matter so we use a combination. There is no natural way to order or rank the patients being tested, and there's no need to do so, because all that matters is the group itself rather than specific individuals.

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Problem 23

Again, order doesn't matter because we arent worried about rank or position. All we care about is the group. We use a combination here as well.

3 0

3 years ago

Evaluate the expression s=5 and t=7. 3st to the second power is what

bija089 [108]

$s=5;\ t=7\\\\Subtitute\ the\ values\ of\ s\ and\ t\ to\ the\ expression\ 3st^2:\\\\3\cdot5\cdot7^2=15\cdot49=735$

7 0

3 years ago

The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa. What is

tensa zangetsu [6.8K]

Answer:

The first quartile of the strengths of this alloy is 9.055 GPa.

Step-by-step explanation:

Problems of normal distributions 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.

The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa.

This means that $\mu = 10, \sigma = 1.4$

What is the first [lower] quartile of the strengths of this alloy?

This is the 100/4 = 25th percentile, which is X when Z has a pvalue of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 10}{1.4}$

$X - 10 = -0.675*1.4$

$X = 9.055$

The first quartile of the strengths of this alloy is 9.055 GPa.

3 0

3 years ago

The probability that any child in a certain class travels to school by bus is 2/5. If 4 children from the class are selected ata

RUDIKE [14]

Hmmm i would have to say 25 percent

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3 years ago

Simplify the expression below<br> 17. 5 + 3 . (8.1 - 2.3)

Free_Kalibri [48]

The expression is 22.4.

8 0

3 years ago