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Aleonysh [2.5K]
3 years ago
9

Please enter the missing number: 4, 8, 14, 22, ?

Mathematics
2 answers:
nevsk [136]3 years ago
7 0
C:32
Because if you look at the numbers it’s going in a pattern from 4 to 8 it’s +4 the. 8 to 14 is +6, 14 to 22 is +8 so the next one would be 22 +10 which is 32
lara31 [8.8K]3 years ago
5 0

Answer:

C) 32

Explanation:

it's adding by 4, 6, 8, and then 10; 22 + 10 = 32, so C will be the answer.

Hope this helps!

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The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with a
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Answer:

1) The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

2) 15.87% of all pregnancies last less than 250 days

2a) 83.5% of pregnancies last between 241 and 286 days

2b) 10.56% of pregnancies last more than 286 days.

2c) 0% of pregnancies last more than 333 days

3) A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

4) The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

5) The most typical 72% of all pregnancies last between 248.72 and 283.28 days.

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

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.

Mean of 266 days and a standard deviation of 16 days.

This means that \mu = 266, \sigma = 16

(1) Using the 68-95-99.7% rule, between what two lengths do the most typical 68% of all pregnancies fall 95%, 99.7%?

68%: within 1 standard deviation of the mean, so 266 - 16 = 250 days to 266 + 16 = 282 days.

95%: within 2 standard deviations of the mean, so 266 - 32 = 234 days to 266 + 32 = 298 days.

99.7%: within 3 standard deviations of the mean, so 266 - 48 = 218 days to 266 + 48 = 314 days.

The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

(2) What percent of all pregnancies last less than 250 days?

The proportion is the p-value of Z when X = 250. So

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

Z = \frac{250 - 266}{16}

Z = -1

Z = -1 has a p-value of 0.1587.

0.1587*100% = 15.87%.

15.87% of all pregnancies last less than 250 days.

(a) What percentage of pregnancies last between 241 and 286 days?

The proportion is the p-value of Z when X = 286 subtracted by the p-value of Z when X = 241. So

X = 286

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

Z = \frac{286 - 266}{16}

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

X = 241

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

Z = \frac{241 - 266}{16}

Z = -1.56

Z = -1.56 has a p-value of 0.0594.

0.8944 - 0.0594 = 0.835*100% = 83.5%

83.5% of pregnancies last between 241 and 286 days.

(b) What percentage of pregnancies last more than 286 days?

1 - 0.8944 = 0.1056*100% = 10.56%.

10.56% of pregnancies last more than 286 days.

(c) What percentage of pregnancies last more than 333 days?

The proportion is 1 subtracted by the p-value of Z = 333. So

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

Z = \frac{333 - 266}{16}

Z = 4.19

Z = 4.19 has a p-value of 1

1 - 1 = 0% of pregnancies last more than 333 days.

(3) What length cuts off the shortest 2.5% of pregnancies?

This is the 2.5th percentile, which is X when Z = -1.96.

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

-1.96 = \frac{X - 266}{16}

X - 266 = -1.96*16

X = 234.6

A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

(4) Find the quartiles for pregnancy length.

First quartile the 25th percentile, which is X when Z = -0.675.

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

-0.675 = \frac{X - 266}{16}

X - 266 = -0.675*16

X = 255.2

Third quartile is the 75th percentile, so X when Z = 0.675.

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

0.675 = \frac{X - 266}{16}

X - 266 = 0.675*16

X = 276.8

The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

(5) Between what two lengths are the most typical 72% of all pregnancies?

Between the 50 - (72/2) = 14th percentile and the 50 + (72/2) = 86th percentile.

14th percentile:

X when Z = -1.08.

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

-1.08 = \frac{X - 266}{16}

X - 266 = -1.08*16

X = 248.72

86th percentile:

X when Z = 1.08.

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

1.08 = \frac{X - 266}{16}

X - 266 = 1.08*16

X = 283.28

The most typical 72% of all pregnancies last between 248.72 and 283.28 days.

7 0
3 years ago
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