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Arturiano [62]
4 years ago
12

Construct a​ 95% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A sample of

25 randomly English majors has a mean test score of 81.5 with a standard deviation of 10.2. A. ​(77.29, 85.71) B. ​(66.35, 69.89) C. ​(56.12, 78.34) D. ​(87.12, 98.32)
Mathematics
1 answer:
blagie [28]4 years ago
3 0

Answer:

reeeeeeeeeeeee

Step-by-step explanation:

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rosijanka [135]

Answer:

Neither A or B

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If a number is added to twice the number the result is -9 what is the number
Licemer1 [7]

Answer:

the number is added to twice number is 99

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3 years ago
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The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo
dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

\to f_X (x) \ \ xe^{-x} \ , \ x>0

For point a:

Moment generating function of X=?

Using formula:

\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx

M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx

integrating the values by parts:

u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0}  -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\  

        = \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\

Therefore, the moment value generating by the function is =\frac{1}{(t-1)^2}

In point b:

E(X^n)=?

Using formula: E(X^n)= M^{n}_{X}(0)

form point (a):

\to M_{X}(t)=\frac{1}{(t-1)^2}

Differentiating the value with respect of t

M'_{X}(t)=\frac{-2}{(t-1)^3}

when t=0

M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\

M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\

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3 years ago
The sum of two numbers is 27 and their product is 50. Find the numbers.?<br><br>​
san4es73 [151]

Answer:

a + b = 27

=> a = 27 - b

ab = 50

So, (27 - b) x b = 50

27b - b^2 = 50

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Solving, b = 25, 2

Hence,

Case 1: b = 25, a = 2

Case 2, b = 2, a = 25

5 0
2 years ago
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How do you do this question?
Maslowich

Answer:

11.736

Step-by-step explanation:

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7 0
3 years ago
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