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3 years ago

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There is a bag filled with 3 blue, 4 red and 5 green marbles. A marble is taken at random from the bag, the colour is noted and

then it is replaced. Another marble is taken at random. What is the probability of getting exactly 1 green?

1 answer:

aliina [53]3 years ago

8 0

Answer:

i think the answer is 9.

Step-by-step explanation:

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181.93 divided by -11.3

jonny [76]

Answer:

-16.1

Step-by-step explanation:

4 0

3 years ago

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An ice cream factory makes 250 quarts of ice cram In 5 hours. How many quarts could be made in 36 hours? What was that rate per

aniked [119]

Answer:1900 in 36 hours and 1200 in 24 hours

Step-by-step explanation:

Find how many they can make in an a hour first. 250 divided by 5. That makes 50. Multiply that by the amount of hours

5 0

3 years ago

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Please help solve by elimination I hope its not blurry Have a nice day and Goodnight

strojnjashka [21]

Answer:

$\displaystyle \textcolor{black}{4.}\:[-3, -5]$

$\displaystyle \textcolor{black}{3.}\:[8, -1]$

$\displaystyle \textcolor{black}{2.}\:[4, -7]$

$\displaystyle \textcolor{black}{1.}\:[3, -2]$

Step-by-step explanation:

When using the Elimination method, you eradicate one pair of variables so they are set to zero. It does not matter which pair is selected:

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-5x - 3y = 30}} \right.$

{2x - 3y = 9

{⅖[−5x - 3y = 30]

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-2x - 1\frac{1}{5}y = 12}} \right. \\ \\ \frac{-4\frac{1}{5}y}{-4\frac{1}{5}} = \frac{21}{-4\frac{1}{5}} \\ \\ \boxed{y = -5, x = -3}$

------------------------------------------------------------------------------------------

$\displaystyle \left \{ {{x - 2y = 10} \atop {x + 3y = 5}} \right.$

{x - 2y = 10

{⅔[x + 3y = 5]

$\displaystyle \left \{ {{x - 2y = 10} \atop {\frac{2}{3}x + 2y = 3\frac{1}{3}}} \right. \\ \\ \frac{1\frac{2}{3}x}{1\frac{2}{3}} = \frac{13\frac{1}{3}}{1\frac{2}{3}} \\ \\ \boxed{x = 8, y = -1}$

_______________________________________________

$\displaystyle \left \{ {{y = -3x + 5} \atop {y = -8x + 25}} \right.$

{y = −3x + 5

{−⅜[y = −8x + 25]

$\displaystyle \left \{ {{y = -3x + 5} \atop {-\frac{3}{8}y = 3x - 9\frac{3}{8}}} \right. \\ \\ \frac{\frac{5}{8}y}{\frac{5}{8}} = \frac{-4\frac{3}{8}}{\frac{5}{8}} \\ \\ \boxed{y = -7, x = 4}$

------------------------------------------------------------------------------------------

$\displaystyle \left \{ {{y = -x + 1} \atop {y = 4x - 14}} \right.$

{y = −x + 1

{¼[y = 4x - 14]

$\displaystyle \left \{ {{y = -x + 1} \atop {\frac{1}{4}y = x - 3\frac{1}{2}}} \right. \\ \\ \frac{1\frac{1}{4}y}{1\frac{1}{4}} = \frac{-2\frac{1}{2}}{1\frac{1}{4}} \\ \\ \boxed{y = -2, x = 3}$

_______________________________________________

I am joyous to assist you at any time.

7 0

2 years ago

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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

What is 51.5% of 845 million

Masja [62]

The answer is 435,175,000.

4 0

3 years ago