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2 years ago

luis writes a five page report with 250 words. at the same rate, how many pages would be needed for 400 words?

Mathematics

1 answer:

Georgia [21]2 years ago

3 0

Answer:

you will need 8 pages for 400 words.

Step-by-step explanation:

250/5=50 (divided 250 with 5 and got 50)

Then I took 400 and divided it by 50 and got 8. (400/50=8)

I hope this helps!

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What is f(x)=2x^2+3x+7 is x=-5

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f(-5) = 42

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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

3 years ago

Can somebody help me on 23 and 24, it’s geometry

Luda [366]

Question 23:

x = 4

DE = 44

Question 24:

x = 25

SE = 28

Step-by-step explanation:

As RS is the perpendicular bisector of DE, it will divide DE in two equal parts DS and SE

Question number 23:

Given

DS = 3x+10

SE = 6x-2

As the two segments are equal:

$DS = SE\\3x+10 = 6x-2$

Subtracting 10 from both sides

$3x+10-10 = 6x-2-10\\3x = 6x-12$

subtracting 6x from both sides

$3x -6x = 6x-6x-12\\-3x = -12$

Dividing both sides by -3

$\frac{-3x}{-3} = \frac{-12}{-3}\\x = 4$

Now

$DS = 3x+10\\= 3(4)+10\\= 12+10\\=22$

And

$SE = 6x-2\\= 6(4)-2\\= 24 - 2\\=22\\DE = DS+SE\\= 22+22\\=44$

Question No 24:

Given

DS = x+3

DE = 56

We know that:

$DS = \frac{1}{2}DE\\x+3 = \frac{56}{2}\\x + 3 = 28\\x = 28-3\\x = 25$

DS = $25+3 = 28$

As DS is 28, SE will also be 28

Hence,

Question 23:

x = 4

DE = 44

Question 24:

x = 25

SE = 28

Keywords: Bisector, Line segment

Learn more about line segments at:

#LearnwithBrainly

4 0

3 years ago

What is the sum of the complex numbers 2 + 3i and 4 + 8i, where i = −1 ?

kotykmax [81]

-1, -4 because 2+3(-1)= -1 and 4+8(-1)= -4

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3 years ago

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