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kodGreya [7K]
3 years ago
5

A baby weighed 7.25 lb at birth at the end of 8 months , the baby weighed 2 1/2 times its birth weight. how many pounds did the

baby weigh at the end of 8 month ?
Mathematics
2 answers:
Sedbober [7]3 years ago
4 0

Answer:

The weight of baby at the end of 8 month is 18.125 lb.

Step-by-step explanation:

Given information:

The weight of baby at birth = 7.25 lb

At the end of 8 months, the baby weighed 2\frac{1}{2} times its birth weight.

We need to find the weight of baby at the end of 8 month.

2\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}

The weight of baby at the end of 8 month is 5/2 times of its initial weight.

\text{Total weight}=7.25\times \frac{5}{2}

\text{Total weight}=18.125

Therefore the weight of baby at the end of 8 month is 18.125 lb.

d1i1m1o1n [39]3 years ago
3 0
Simply multiply the baby's original weight, which is 7.25 lbs, by 2.5

7.25lbs x 2.5 = 18.125 lbs

The baby weighted 18.125 pounds at the end of 8 months.

Hope this helps and May the Force Be With You!

-
Jabba
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Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
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Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

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3 years ago
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