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3 years ago

1. Given: 3(x - 6) + 10 = 31 Prove: x = 13

Mathematics

1 answer:

Dima020 [189]3 years ago

8 0

Answer:

<h3> The Answer is proved.</h3>

Step-by-step explanation:

<h3> If you like this answer then mark this answer as BRAINLIEST answer....</h3>

Thank you ☺️☺️

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What is the Mean Average deviation from the Mean?<br> Numbers: 6, 2, 3, 3, 2, 2

Irina-Kira [14]

Answer:

2.42

Step-by-step explanation:

standard deviation= sqrt(variance)

Variance=E[X2] - E[X]2

E[X] = sum(X)/n=35/9, so E[X]2 = 1225/81

E[X2] =sum(X2)/n = 189/9 = 170/81

Variance= 476/81

Standard deviation = sqrt(476/81) ~2.42

6 0

3 years ago

1/2,1/3,1/8 find the sum

Zarrin [17]

Answer:

23/24

Step-by-step explanation:

To find the sum you must ad.

Fractions can only be added and subtracted when the denominators (the bottom numbers) are the same.

24 is the smallest number that 2, 3 and 8 go into.

1/2 = 12/24, 1/3 = 8/24, 1/8 = 3/24

$\frac{12}{24} + \frac{8}{24} + \frac{3}{24} = \frac{23}{24}$

23/24

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4 years ago

Please hurry please & thank you

Mama L [17]

All rhombuses have 2 pairs of parallel sides.

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3 years ago

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

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.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

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