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Shkiper50 [21]
3 years ago
14

The latitude value of Toronto is 43.70 degrees, and the longitude value is 79.40 degrees. The latitude value of Melbourne is -37

.81 degrees, and the longitude value is 144.96 degrees. The two cities are degrees apart in latitude. The two cities are degrees apart in longitude.
Mathematics
1 answer:
Aneli [31]3 years ago
3 0
I'm hazarding a guess it's 224.36
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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
3 years ago
Please need help on this question 20 points
notsponge [240]

Answer:C

Step-by-step explanation:

3 0
3 years ago
In a survey, 150 people were asked if they smoke and whether or not they exercise regularly (more than twice a week). The result
s2008m [1.1K]
(a) 28%

(b) 24%

When calculating the total amount of people it come out to 150.

According to the table 42 people smoke. 42/150 is 28%

According to the table 36 people exercise regularly. 36/150 is 24%
7 0
3 years ago
Which pairs of quadrilaterals can be shown to be congruent using rigid motions?
OverLord2011 [107]
Answers:
True
Not True
Not True
True
6 0
3 years ago
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Erik and Nita are playing a game with numbers. In the game, they each think of a random number from zero to 20. If the differenc
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Let X is the random number Erik thinks of, and Y is the random number Nita thinks of.
Both X and Y are in the range from 0 to 20.
<span>X<=20
Y<=20
If the difference between their two numbers is less than 10, then Erik wins.
The difference between the two numbers can be written X-Y, or Y-X depending on which number (X or Y) is greater. But we do not know that. In order not to get negative value, we calculate absolute value of X-Y,  written |X-Y| which will give positive value whether X is greater than Y or not.
If |X-Y|<10 Erik wins.
</span><span>If the difference between their two numbers is greater than 10, then Nita wins. 
</span><span>If |X-Y|>10 Nita Wins

</span>
6 0
3 years ago
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