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For a certain population of sea turtles, 18 percent are longer than 6.5 feet. A random sample of 90 sea turtles will be selected

nika2105 [10]

Answer:

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

Step-by-step explanation:

Central Limit Theorem:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

18 percent are longer than 6.5 feet.

This means that $p = 0.18$

A random sample of 90 sea turtles

This means that $n = 90$

What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.18*0.82}{90}} = 0.0405$

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

8 0

3 years ago

What is the best approximation of the circumference of a circular pizza pie whose radius is 6 inches?

charle [14.2K]

37.7in i may be wrong but this should help you

6 0

4 years ago

Read 2 more answers

Frank age is one fourth of Dana’s age.

AnnZ [28]

Dana is 16 years oldNeil is 6 years oldFrank is 4 years old

Hope this helps! If you like my answer give me a good vote or brainliest answer!

8 0

3 years ago

Harry read 1/3 of a book on Saturday and three force of the remainder on Sunday. If he still had 75 pages left how many pages we

DaniilM [7]

900

You work backwards;
Since he read 3/4 of the remainder that would be 75 x 4 because 75 was left after 3/4. You would get 300. Then, you would do 300 / (1/3), this is getting the final thing, equals 900.
Hope this helps!

4 0

2 years ago

The water height of a pool is determined by 8g2 + 3g - 4, the rate that the pool is filled, and 9g2 - 2g - 5, the rate that wate

Juliette [100K]

Part A

From the question, we know that

Rate at which pool is filled = 8g² + 3g - 4

Rate at which water leaves the pool = 9g² - 2g - 5

Hence, height of the pool would be = Rate at which pool is filled - Rate at which water leaves the pool

⇒ Height of the pool = 8g² + 3g - 4 - (9g² - 2g - 5)

⇒ Height of the pool = -g² + 5g + 1

Part B

Let water height be H

H (g) = -g² + 5g + 1

When g=1, H(1) = -(1)² + 5 (1) + 1 = 5 units

When g=2, H(2) = -(2)² + 5 (2) + 1 = -4+11 = 7 units

When g=3, H(3) = -(3)² + 5 (3) + 1 = -9+16 = 7 units

When g=4, H(4) = -(4)² + 5 (4) + 1 = -16+21 = 5 units

Part C

Now, we need to determine the value for g for which height would be maximum.

Looking at the expression that determines height, -g² + 5g + 1, we see that the coefficient of g² is negative. Hence the equation would represent a downward facing parabola, which means that the function H(g) will have a maxima point.

To find out the maxima point, differentiate H(g) with respect to g, and equate the resulting expression to zero.

$\frac{d(H(g)}{dg}$ = -2g + 5 = 0

⇒ 2g = 5

⇒ g = 2.5

So at g = 2.5, the height of the water in the pool is maximum. [Note: nearest tenth means rounding till the first decimal point, hence the answer is g=2.5]

8 0

3 years ago

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