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mrs_skeptik [129]
3 years ago
10

Can someone explain to me how to do number 5. ✌️

Mathematics
1 answer:
Dahasolnce [82]3 years ago
5 0
Fist u have to calculate how much it takes to reach the certain number for instance it takes 30 per gallon because 30 times 1 is 30. 30 times 2 is 60 and so on
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1.) A number has 7 tens and 3 more ones than tens. What<br> is the number?<br> be
Lubov Fominskaja [6]

Answer:

80

Step-by-step explanation:

hope it helps . .. ..,......

5 0
3 years ago
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38 divided by 3576 long division<br><img src="https://tex.z-dn.net/?f=38%20%5Csqrt%7B3576%7D%20" id="TexFormula1" title="38 \sqr
Scilla [17]

Answer:

you will get a decimal answer which is 94.10526316

Step-by-step explanation:

8 0
3 years ago
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
Five friedns share three pizzas that cost 12.75 how much do each of them pay
Aliun [14]
They each would pay 2.55

12.75 divided by 5 = 2.55
5 0
3 years ago
Help me out with this
ArbitrLikvidat [17]

Answer:

What is 2x + 5 (y-1) when x=8 and y=5

Step-by-step explanation:

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2 years ago
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