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Mazyrski [523]
3 years ago
12

A student working on a report about mathematicians decides to find the 98% confidence interval for the difference in mean age at

the time of math discovery for Greek mathematicians versus Egyptian mathematicians. The student finds the ages at the time of math discovery for members of both groups, which include all Greek and Egyptian mathematicians, and uses a calculator to determine the 98% confidence interval based on the t distribution. Why is this procedure not appropriate in this context? (4 points)
Mathematics
1 answer:
balandron [24]3 years ago
6 0

Answer:

The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.

Step-by-step explanation:

A student working on a report about mathematicians decides to find the 98% confidence interval for the difference in mean age at the time of math discovery for Greek mathematicians versus Egyptian mathematicians. The student finds the ages at the time of math discovery for members of both groups, which include all Greek and Egyptian mathematicians, and uses a calculator to determine the 98% confidence interval based on the t distribution, <u>the entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.</u> There is no as such explanation of this scenario, its just the answer mentioned above.

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The bill at a restaurant came to $136.40.The customer decided to leave a 15% tip.What was the total bill including tip.
hjlf

Answer:

$156.86

Step-by-step explanation:

To calculate tip, you need to multiply the total bill by the percent tip you are going to leave. In this case the bill is $136.40, and the customer is going to leave a 15% tip. So, you would want to move the decimal place over to the left by 1, which you should get 13.64 as your number. Then divide it in half, which should give you 6.82. Now add 13.64 and 6.82 with the original bill price. If you don't add the 13.64 you would've forgotten the 10% part of the tip. At the end you should get $156.86 as your final bill.

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3 years ago
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
What is an equation of the line that passes through the point ( 6 , − 2 ) (6,−2) and is perpendicular to the line 6 x + y = 2 6x
Diano4ka-milaya [45]

Answer:

An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

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Step-by-step explanation:

We know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept.

Given the line

6x+y=2

Simplifying the equation to write into the  slope-intercept form

y = -6x+2

So, the slope = -6

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

Thus, the slope of the perpendicular line will be: -1/-6 = 1/6

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be

y-y_1=m\left(x-x_1\right)

substituting the values m = 1/6 and the point (6, -2)

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y+2=\frac{1}{6}\left(x-6\right)

subtract 2 from both sides

y+2-2=\frac{1}{6}\left(x-6\right)-2

y=\frac{1}{6}x-3

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

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Whats the exponential form for -1/6×-1/6×-1/6×-1/6=?
KengaRu [80]
I hope this helps you

4 0
3 years ago
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