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Keith_Richards [23]
3 years ago
14

13=2f+5 f=two step equation

Mathematics
1 answer:
stepan [7]3 years ago
6 0
2f+5=13  subtract 5 from both sides

2f=8  divide both sides by 2

f=4
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Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u=207 and standard d
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Answer:

a) 36.32% probability that a randomly selected pregnancy lasts less than 201 days.

b) 3.22% probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less.

c) 1.58% probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less

Step-by-step explanation:

To solve this problem, we have to know the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit theorem

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 207, \sigma = 17

(a) what is the probability that a randomly selected pregnancy lasts less than 201 days?

Pvalue of Z when X = 201.

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

Z = \frac{201 - 207}{17}

Z = -0.35

Z = -0.35 has a pvalue of 0.3632.

So 36.32% probability that a randomly selected pregnancy lasts less than 201 days.

(b) what is the probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less?

Now n = 16, s = \frac{17}{\sqrt{16}} = 3.25

Again the pvalue of Z when X = 201

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

Due to the Central Limit Theorem

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

Z = \frac{201 - 207}{3.25}

Z = -1.85

Z = -1.85 has a pvalue of 0.0322.

So 3.22% probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less.

(c) what is the probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less?

Now n = 16, s = \frac{17}{\sqrt{37}} = 2.79

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

Z = \frac{201 - 207}{2.79}

Z = -2.15

Z = -2.15 has a pvalue of 0.0158.

1.58% probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less

8 0
4 years ago
The base of a right prism is a square with edge 4 . The volume is 128 . The height is ?
ollegr [7]
Volume = (length) x (width) x (height)

Height = (Volume) / (length x width) = 128 / (4 x 4) = 128/16 = <u>8</u> .
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Answer:

The points form a curve

Step-by-step explanation:

Jk the points are in a straight line

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Answer:

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