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soldi70 [24.7K]
3 years ago
9

A baby gains about 15 pounds each month. About how much weight will the baby have gained after 6 months? a. About 7.5 pounds b.

About 8.5 pounds C. About 9 pounds d. About 10.5 pounds​
Mathematics
2 answers:
Molodets [167]3 years ago
7 0

Answer:

I'm going to assume you meant 1.5 pounds and not 15 pounds each month.

Step-by-step explanation:

1.5*6=9

C

Svetach [21]3 years ago
5 0

Answer:

around 90 pounds

Step-by-step explanation:

15×6

or 15+15+15+15+15+15 if multiplying is not your thing

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At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
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0.0869 = 8.69% probability of getting more than 61% green balls.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

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

By the Central Limit Theorem

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

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

8 0
3 years ago
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