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docker41 [41]
3 years ago
15

How to do the problem

Mathematics
1 answer:
gavmur [86]3 years ago
5 0

Distribute the outside number (the 2 fractions in this case) to the numbers in the parenthesis. Multiply them and then solve for the variable. To solve for variable just add or subtract the constant (add if negative subtract if positive). Then divide each side with the number paired with the variable.

Just ask if you want me to edit answers in

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3 0
2 years ago
An amusement park sells 5 bottles of water for every
77julia77 [94]

Answer:

8 bottles of juice

3 0
3 years ago
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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
2 years ago
11 more than a number q is negative 15 ; Solve for the variable.
Cloud [144]

Answer:

q = -26

Step-by-step explanation:

Set up an equation where q is the unknown number:

q + 11 = -15

Solve for q by subtracting 11 from both sides:

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4 0
2 years ago
A grocery store sells a bag of 6 oranges for $2.34. If Mav spent $1.95 on oranges, how many did she buy?
poizon [28]

Answer:

5 oranges

Step-by-step explanation:

to solve this problem we need to make a cross-multiplication equation

2.34 is to 6 as 1.95 is to ?

? is the number of oranges that are needed to be found that were bought

2.34/6 = 1.95/?

Now we cross multiply 6 and 1.95 = 11.7

11.7/2.34 = 5

Give brainliest, please!
hope this helps :)

5 0
2 years ago
Read 2 more answers
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