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Zielflug [23.3K]
3 years ago
5

The price of cappuccinos is $5.00 each. The price of pistachios is $1.00 per pound. If Curtiss has income of $32.00 and purchase

s 5.00 cappuccinos, how many pounds of pistachios can he buy
Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

curtiss can buy 7 pounds of pistachios

Step-by-step explanation:

5 cappuccinos x $5 = 25, 32 - 25=7

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(1, 5), (3, 19) (-1, -9)

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Emma has 7 bottles: 3 bottles that contain juice, 2 bottles
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Answer: Out of 7 bottles, 2 bottles contain soda. This means that 2 bottles= 29% out of 100%. 7/2 (7 divided by 2) = 0.285. 0.285 can be rounded up to 0.29, meaning that the approximate percentage of the bottles that contain soda is 29%.

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. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

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}}

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.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

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3 years ago
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Step-by-step explanation:

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3 years ago
I need help with this please
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Answer:32

Step-by-step explanation:

5 0
3 years ago
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