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Vera_Pavlovna [14]
3 years ago
5

Given a mean of 150 and a standard deviation of 25:

Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0
We have to define an interval about the mean that contains 75% of the values. This means half of the values will lie above the mean and half of the values lie below the mean.

So, 37.5% of the values will lie above the mean and 37.5% of the values lie below the mean.

In a Z-table, mean is located at the center of the data. So the position of the mean is at 50% of the data. So the position of point 37.5% above the mean will be located at 50 + 37.5 = 87.5% of the overall data

Similarly position of the point 37.5% below the mean will be located at 
50 - 37.5% = 12.5% of the overall data

From the z table, we can find the z value for both these points. 12.5% converted to z score is -1.15 and 87.5% converted to z score is 1.15.

Using these z scores, we can find the values which contain 75% of the values about the mean.

z score of -1.15 means 1.15 standard deviations below the mean. So this value comes out to be:

150 - 1.15(25) = 121.25

z score of 1.15 means 1.15 standard deviations above the mean. So this value comes out to be:

150 + 1.15(25) = 178.75

So, the interval from 121.25 to 178.75 contains the 75% of the data values. 
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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
3 years ago
Last one - Thank you an advance :)
Ivan
Answers:
a. Given
b. Transitive property of congruence
c. Vertical angles are congruent
d. Transitive property of congruence

Let me know if you need any clarification as to how I got those answers. They should be self-explanatory but I'm happy to clarify further if needed. 

5 0
3 years ago
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What is 22 nickels equal to?
Viktor [21]
$1.10 is your answer
4 0
3 years ago
Name the variables : r + y - 3
Naddika [18.5K]

Answer:

3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
PLS HELP!
wolverine [178]

Given:

Alexis has a rectangular backyard that is 50 yards by 55 yards.

She wants to build a fence that stretches diagonally from one corner to the opposite corner.

To find:

The length of the fencing she needs.

Solution:

We have,

Length = 50 yards

Width = 55 yards

We know that, the diagonal of a rectangle is

Diagonal=\sqrt{length^2+width^2}

Diagonal=\sqrt{(50)^2+(55)^2}

Diagonal=\sqrt{2500+3025}

Diagonal=\sqrt{5525
}

On further simplification, we get

Diagonal=74.3303437

Diagonal\approx 74.3

Therefore, the length of the required fencing is 74.3 yards.

8 0
3 years ago
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