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PolarNik [594]
3 years ago
14

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?
Mathematics
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
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