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2 years ago

The variable is Z is inversely proportional to X. When X is 6, Z has the value 0.5. What is the value of Z when X = 10

1 answer:

6 0

Given that z is inversely proportional to x, then:
zα1/x
hence;
z=k/x
where k is the constant of proportionality;
this implies that:
k=xz
when x=6,z=0.5
thus;
k=0.5*6
k=3
hence;
z=3/x
when x=10,
z=3/10
the answer is 3/10

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Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, 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 establishes 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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

2 years ago

Find the sum of the numbers as the product of their GCF and another sum. 35 + 49

AlladinOne [14]

Answer:

84 is what you get when you add them

14 is what you get when you factor their GCF then add

hope this helps plz mark brainliest

3 0

2 years ago

How do I solve it because it’s really difficult

dybincka [34]

The solution of the equation is x= 3 and y = 2

Step-by-step explanation:

Given,

-3x-4y = -17 and

-x-3y = -9

The system of equation can be rewritten as

3x+4y = 17 ------eq 1 and

x+3y = 9 ------ eq 2

To solve for x and y

Multiplying eq 2 by 3 we get,

3x + 9y = 27

or, 3x = 27-9y

Putting this value of x into eq 1 we get,

27-9y +4y = 17

or, -5y = 17-27

or, -5y = -10

or, y = 2

Now put y=2 in eq 2 we get,

x = 9 - 3(2)

= 3

Hence the solution is x = 3 and y = 2

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2 years ago

Retha Davidson works for a word processing center keying straight copy.

Nataliya [291]

Using proportions, it is found that:

In an 8-hour day, she keys 48 pages.

For a day, her gross earnings are of $72.

For a week, her gross earnings are of 72 x 7 = $504.

<h3>Proportions:</h3>

This question is solved using the proportions given, which are:

Payment of $1.50 per page of 30 lines.

Retha keys 6 pages per hour.

Hence, in an 8-hour day, she keys 6 x 8 = 48 pages.

Since she earns $1.50 per page, for a day, her gross earnings are of 48 x 1.5 = $72.

Considering a week has 7 days, for a week, her gross earnings are of 72 x 7 = $504.

To learn more about proportions, you can take a look at brainly.com/question/24372153

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1 year ago