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11Alexandr11 [23.1K]
4 years ago
7

Elly and Drew work together to collect data to estimate the percentage of their classmates who own a particular brand of shoe. U

sing the same data, Elly will construct a 90 percent confidence interval and Drew will construct a 99 percent confidence interval. Which of the following statements is true?a. The midpoint of Elly's interval will be greater than the midpoint of Drew's interval. b. The midpoint of Elly's interval will be less than the midpoint of Drew's interval. c. The width of Elly's interval will be greater than the width of Drew's interval. d. The width of Elly's interval will be less than the width of Drew's interval. e. The width of Elly's interval will be equal to the width of Drew's interval
Mathematics
1 answer:
ladessa [460]4 years ago
4 0

Answer:

d. The width of Elly's interval will be less than the width of Drew's interval.

Step-by-step explanation:

The confidence level and the width of the confidence interval are direct proportional. This means that a confidence interval with a higher confidence level has a higher width.

For example, a 99 percent confidence interval is wider than a 90 percent confidence interval.

The midpoint of the confidence interval is the mean of the population, no matter the confidence level.

In this problem, we have that:

Elly: 90 percent CI

Drew: 99 percent CI

The correct answer is:

d. The width of Elly's interval will be less than the width of Drew's interval.

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Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

6 0
2 years ago
Read 2 more answers
Sandra ran 7/12 of a mile lamar ran 3/4 a mile who ran farther explain pls help
LekaFEV [45]
Sandra ran 7/12 of a mile. Lamar ran 3/4 of a mile. Who ran further?
Lamar ran further.
3/4=(3*3)/(4*3) which equals 9/12.
Based off calculations we also see that 9/12>7/12.
Which then leaves us too 3/4>7/12.
3 0
3 years ago
Read 2 more answers
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 64 women are
shtirl [24]

Answer:

0.3569 is the probability that they have a mean pregnancy between 266 days and 268 days.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  268 days

Standard Deviation, σ =  15 days

We are given that the distribution of lengths of pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{64}} = \frac{15}{8}

P(pregnancy between 266 days and 268 days)

P(266 \leq x \leq 268) = P(\displaystyle\frac{266 - 268}{\frac{15}{8}} \leq z \leq \displaystyle\frac{268-268}{\frac{15}{8}}) = P(-1.0667 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1.067)\\= 0.5000 - 0.1431 = 0.3569 = 35.69\%

P(266 \leq x \leq 268) = 35.69\%

6 0
3 years ago
Solve for a.
prohojiy [21]
The correct answer is -10
4 0
3 years ago
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Plzzzzz help just tell me a b c d for both of them ill give anyone brainliest too
Hoochie [10]

Question 1 (The wood question)

Answer: 14 7/8

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I hope this helps!

4 0
3 years ago
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