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3 years ago

the diameter of the earth at the equator is about 8,000 mi. Based on this figure ,how far is it exactly around the earth

2 answers:

8 0

Basically, we want to find the circumfernece of the earth
C=pid
d=diameter
pi=pi (aprox pi=3.1419256)
C=3.14159256*8000
C=25132.74048 feet

geniusboy [140]3 years ago

8 0

C = 2 π r
C = 8,000 π or 25,133 miles

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Choose the expression that represents the indicated composition. f(x) = 10x g(x) = x −3 (f º g)(x) = ? 10x – 30 10x – 3 10x2 – 3

8090 [49]

(fºg)(x)=f(g(x))=f(x-3)=10(x-3)=10x-30

3 0

3 years ago

Read 2 more answers

The weight of an organ in adult males has a bell-shaped distribution with a mean of 320 grams and a standard deviation of 45 gr

irina1246 [14]

Answer:

a) 99.7%

b) 68%

c) 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

d) 81.5% of organs weighs between 290 grams and 365 grams.

Step-by-step explanation:

A) 99.7% of organs will be between 3 standard deviation from the mean.

$320 - 3 \times 45 = 185$

$320 + 3 \times 45 = 455$

So 99.7% of organs will be between 185 and 455.

B) 275 grams and 365 grams are 1 standard deviation from the mean.

From empirical rule about 68% data falls within 1 standard deviation from the mean.

So 68% of organs weighs btwn 275 grams and 365 grams.

C) Since 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

D) 230 is 2 standard deviation below the mean and 365 is one standard deviation above the mean.

According to the empirical rule, 95% of the observation lies within 2 standard deviations of the mean. Therefore 5% lies outside 2 standard deviations of the mean.

So 95%/2 = 47.5% of organs weighs between the mean and 230 grams.

According to the empirical rule about 68% data falls within one standard deviation from the mean.

so, 68%/2 = 34% of organs weighs between the mean and 365 grams.

So total = 47.5% + 34% = 81.5% of organs weighs between 290 grams and 365 grams.

7 0

3 years ago

A wrestler competes in 25 matches. One of those matches, he wins 17. What percent of the matches did the wrestler win?

PSYCHO15rus [73]

68 it think

Step-by-step explanation:

6 0

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

PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER

drek231 [11]

Answer:

Step-by-step explanation:

The line of best fit is a linear function that is closest to all the points and shows the trend in the data. Since the data is going up ultimately, the slope of the line will be positive. Only B and C are options.

To find the best fit, draw a line through the center of the most points. This line will have as its y-intercept a value around 12 and a gentle slope. This means the answer choice is C.

3 0

4 years ago