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attashe74 [19]
3 years ago
15

Please, does anyone know?

Mathematics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

-16, 5, 20, -4, 83

Step-by-step explanation:

You replace x with the given numbers on the chart, then solve.

For example: 3(-7)+5

-21+5= -16

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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
At a point on the ground 46 feet from the foot of a tree, the angle of elevation of the top of the tree is 68 degrees. What is t
Mandarinka [93]

Answer:

114 ft

Step-by-step explanation:

Imagine or construct a right triangle with the 46 ft leg lying on the ground.  This is the "adjacent side" of the triangle; it lies immediately adjacent to the 68 degree angle.  The side opposite this angle is h, the height of the tree.

The tangent function includes angle, opp side and adj side:

tan 68 degrees = opp / adj = h / (46 ft), and so:

(46 ft)*tan (68 degrees) = opp = h

Then the height of the tree is h = (46 ft)(2.47) = 114 ft

6 0
3 years ago
7¼× 8/9×3/4<br>find the sum ​
larisa86 [58]

Answer:

Step-by-step explanation:

7\frac{1}{4}*\frac{8}{9}*\frac{3}{4}=\frac{29}{7}*\frac{8}{9}*\frac{3}{4}\\\\=\frac{29*8*3}{7*9*3}\\\\=\frac{29*2*1}{7*3*1}\\\\=\frac{58}{21}\\\\=2\frac{16}{21}\\

7 0
2 years ago
26. Julie says that sin A/cos A = tan A and sin B / cos B = tan B. Use the triangle shown to prove
PSYCHO15rus [73]

Julie is correct because:

\frac{\sin A}{\tan A}=\frac{a/c}{b/c}=a/b=\tan A \\ \\ \frac{\sin B}{\cos B}=\frac{b/c}{a/c}=b/a =\tan B

5 0
1 year ago
(5x + 2)<br> 72°<br> I need help pls fast asap
Misha Larkins [42]

69818247034ue8e83884647

8 0
2 years ago
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