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4 years ago

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Angie and Kenny play online video games. Angie buys 2 software packages and 3 months of gameplay . Kenny buys 2 software package

s and 2 months of gameplay . Each software package costs $35. Of their total cost is $210 , what is the cost of one month of gameplay?

1 answer:

sveta [45]4 years ago

7 0

I believe not 100% on this one that the answer is $14 for the cost of one month of gameplay. How I came up with that is answer is that if you add all the softwares they brought which is 4 and times that by $35. $35*4= 140. Then subtract 210-140=70. Then you divide that by month of gameplay which will be 70/5=14. That's how i came up with $14

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Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa

Debora [2.8K]

Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 8.7% and standard deviation 20.2%.

This means that $\mu = 8.7, \sigma = 20.2$

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This means that $n = 40, s = \frac{20.2}{\sqrt{40}}$

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This is 1 subtracted by the pvalue of Z when X = 13. So

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This is the pvalue of Z when X = 8. So

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

$Z = \frac{8 - 8.7}{\frac{20.2}{\sqrt{40}}}$

$Z = -0.22$

$Z = -0.22$ has a pvalue of 0.4129

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