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xxTIMURxx [149]
3 years ago
15

Sue knows that 2.54 cm is equal to 1 in.She has a piece of fabric that is 25cm.Which proportion will NOT help her find the lengt

h of the fabric in inches?

Mathematics
1 answer:
maksim [4K]3 years ago
5 0
B is correct because you wouldn't get the same answer as the other proportions if you solved it
You might be interested in
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
3 years ago
Distrubutive property 4(c+d+9)
ivann1987 [24]

Answer:

4c+4d+36

Step-by-step explanation:

multiply each separate variable inside the parenthesis by the number outside of parenthesis.

7 0
3 years ago
Explain step by step how to use the horizontal method the multiple the two polynomials
MatroZZZ [7]

Answer:

Step-by-step explanation:

Multiply each term of the first polynomial with   the second polynomial. Then combine the like terms.

(3a<em>² + 5a - 2)* (5a² -3a + 4)</em>

<em> = 3a² *(5a² -3a + 4)  + 5a*(5a² -3a + 4)  - 2*(5a² -3a + 4)</em>

<em>=3a²*5a² - 3a*3a² + 4*3a² +  5a*5a² - 3a*5a  + 4*5a + 5a²*(-2) - 3a*(-2) + 4*(-2)</em>

<em>=15a⁴ - 9a³  + 12a² + 25a³ - 15a² + 20a - 10a² + 6a - 8</em>

<em>= 15a⁴ </em><u><em>- 9a³ + 25a³</em></u><em>   +</em><u><em> 12a² - 15a² - 10a²</em></u><em>   +</em><u><em> 20a +6a </em></u><em>- 8</em>

<em>= 15a⁴ + 16a³  - 13a² +26a  - 8</em>

4 0
3 years ago
Given: AB with coordinates of AC-3, -1) and B(2, 1)
andrezito [222]

Answer:

up 2, right 3

Step-by-step explanation:

up 2, to the right 3.

8 0
3 years ago
Mrs. Little would like to set up each station to be able to make 64 pancakes. How many teaspoons of salt should be measured out
dusya [7]

Answer:

48

Step-by-step explanation:

all you do is 64 times 3/4

5 0
2 years ago
Read 2 more answers
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