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ahrayia [7]
3 years ago
10

Do those two problems have the same answer?

Mathematics
1 answer:
GaryK [48]3 years ago
7 0
Hope this helps that is 64x

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Need help on this ASAPPPPP
lord [1]

Answer: B

Step-by-step explanation:

Over to the right 5 and down 2

3 0
3 years ago
Read 2 more answers
Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
3 years ago
The Symphony
lord [1]
Answer is 12.50. u can comment on my answer if you would like to see how that is the correct answer :)
5 0
3 years ago
A 10-foot ladder is leaning against a tree. The bottom of the ladder is 4 feet away from the bottom of the tree. Approximately h
strojnjashka [21]

Answer:

C

Step-by-step explanation:

If we were to draw a horizontal line from the bottom of the ladder to the bottom of the tree and then draw a vertical line from the bottom of the tree to the top of the ladder, we'd get a right triangle with legs as the distance between the bottom of the tree and the bottom of the ladder and the height of the ladder, and the hypotenuse is the length.

Here, we know the hypotenuse is 10 feet and that the bottom of the ladder is 4 feet away from the bottom of the tree, so use the Pythagorean Theorem to find the height:

h = \sqrt{10^2-4^2} =\sqrt{84} ≈ 9.2 feet

The answer is C.

7 0
3 years ago
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What is the volume of a rectangular prism with a length of 8 1/2 cm, width of 9 1/3 cm and a height of 12 2/5 cm?
AlekseyPX
I believe it is 864, I'm not 100% sure. But I think it is.
8 0
3 years ago
Read 2 more answers
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