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slega [8]
3 years ago
7

Is 5 a solution to the equation 2x^2 + 10 = 60 yes or no

Mathematics
2 answers:
cricket20 [7]3 years ago
8 0
2(5)^2+10
2(25)+10
50+10
60
Yes
notka56 [123]3 years ago
7 0
Hello : 
yes : 2(5)² +10 = 60
50+10=60... ( right ) 

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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!

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3 years ago
When rounded to the nearest hundred thousand thousand will result in 2,400,000
xenn [34]
This question confused me. I don't know what u are asking
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Write the equation of the line shown in the graph. Please help.
jeka57 [31]

Answer:

y = 3

Step-by-step explanation:

The y-coordinate of all points on the graph is 3.

The equation is y = 3

3 0
3 years ago
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The price of a tv was decreased by 20% to £1440. What was the price before the decrease?
grigory [225]

Answer:

€7200

Step-by-step explanation:

20%---------1440

100%----------y

y= 100×1440/20

y= 144000/20

y= €7200

price before decrease= €7200

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2 years ago
on monday the bookstore sold 75 books. on tuesday the bookstore sold 125 books. the bookstore must sell 500 book by friday how m
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There were 200 books sold. so, they need to sell 300 more to have 500 total books sold.
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