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castortr0y [4]
3 years ago
8

Scores on the GRE (Graduate Record Examination) are normally distributed with a mean of 513 and a standard deviation of 82. Use

the 68-95-99.7 Rule to find the percentage of people taking the test who score between 513 and 759. The percentage of people taking the test who score between 513 and 759 is %
Mathematics
1 answer:
Trava [24]3 years ago
4 0

Answer:

Note that we need calculate the scores between the mean (\mu=513) and the mean plus 3 times the standard deviation. It is 3 times since the distance between the values given (513 and 759) divided by the standard deviation is 3 (\frac{759-513}{82}=3)

So the rule say that 99.7% of data is between \mu - 3\sigma and \mu + 3\sigma, as it is a normal distribution half of 99.7% is between \mu and \mu+3\sigma. Hence 49.85% of people score between 513 and 759.

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First of all we will divide 8 by 2 to get the number of full practices in the morning, which is 4. Next divide 7 by 2 to get the number of practices in the afternoon. Although there is a remainder, that remainder is not a full practice session, so we only count the whole numbers which is 3. 
4 + 3 = 7 full practice sessions.
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2 years ago
Simplify the expression where possible.(-4 x^2 )^2
Oduvanchick [21]

Answer:

16x^{4}

Step-by-step explanation:

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(-4x^{2})^{2} = (-4)^{2} *(x^{2})^{2}\\\\\\=16 * x^{2*2}\\\\=16x^{4}

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3 years ago
Ryan is looking for the sum of the cube of a number, n, and 16 What is the sum Ryan is looking for if n=2?
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Given that n=2

And

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6 0
3 years ago
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5 0
3 years ago
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The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

8 0
3 years ago
Read 2 more answers
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