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klemol [59]
3 years ago
13

Jeremy is having his basketball team over for an end-of-season party. He has $80 to spend on pizza and soda. Pizzas cost $13.99

each and cases of soda cost $4.49 each. Write an inequality to represent this scenario. Make sure to define your variables.
Mathematics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

13.99p + 4.49s ≤ 80

Step-by-step explanation:

Pizza is p and soda is s.  If he has only $80 to spend on a combination of both, he cannot obviously spend more than that.  Therefore, the inequality sign is "less than or equal to".  He can spend 80, but he can also spend less than 80.  He cannot spend anything over that, since he doesn't have it to spend!

If pizzas are represented by p, and we know the cost of each one is 13.99, we represent one pizza as 13.99p; if the cost of sida is 4.49 per case, we represent one case as 4.49s.  

The sum of these cannot exceed 80, so the inequality, then, is:

13.99p + 4.49s ≤ 80

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Answer:

0.5 = 50% probability a value selected at random from this distribution is greater than 23

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 23, \sigma = 7

What is the probability a value selected at random from this distribution is greater than 23?

This is 1 subtracted by the pvalue of Z when X = 23. So

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

Z = \frac{23 - 23}{7}

Z = 0

Z = 0 has a pvalue of 0.5

0.5 = 50% probability a value selected at random from this distribution is greater than 23

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