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Valentin [98]
3 years ago
10

Solve: 4(7x-11) +4 = 96 M Insert picture of work​

Mathematics
1 answer:
Svet_ta [14]3 years ago
6 0

Answer:

x = 4 \frac{6}{7}

Step-by-step explanation:

4(7x - 11) + 4 = 96 \\  \\ 28x - 44 + 4 = 96 \\  \\ 28x - 40 = 96 \\  \\ 28x = 96 + 40 \\  \\ 28x = 136 \\  \\ x =  \frac{136}{28}  \\  \\ x =  \frac{34}{7}  \\  \\ x = 4 \frac{6}{7}

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What is another way to show 2 x 4/6
DochEvi [55]

Answer:

2 \frac{4}{6}

I think this is it?? very sorry if it's incorrect

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2 years ago
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Which is better, a really good desert, or a really yummy dinner? What is your favorite?
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Both cause u get some nice dinner then right after u get some nice desert. I just like food so I like anything
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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
The formula for the future value V (in dollars) of an investment earning simple interest is V=p+prt, where p (in dollars) is the
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Answer:

A. p=\dfrac{V}{1+rt}

B. $2,307.69

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You are given the formula

V=p+prt,

where V = investment earning simple interest

p = principal,

r = interest rate

t = time

So,

A. V=p(1+rt)\\ \\p=\dfrac{V}{1+rt}

B. r = 0.06 (or 6% as percent)

V = $3,000

t =5

so,

3,000=p+p\cdot 0.06\cdot 5\\ \\3,000=p+0.3p\\ \\1.3p=3,000\\ \\p=\dfrac{3,000}{1.3}\approx 2,307.69

6 0
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Answer:

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