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Viktor [21]
2 years ago
5

The consecutive odd integers have a sum of 36. find the integers

Mathematics
2 answers:
svet-max [94.6K]2 years ago
5 0
<h3>Answer:  17 and 19</h3>

==================================================

Work Shown:

x = some integer

2x = even integer

2x+1 = odd integer

(2x+1)+2 = 2x+3 = odd integer just after 2x+1

The integers 2x+1 and 2x+3 add to 36

(2x+1)+(2x+3) = 36

2x+2x+1+3 = 36

4x+4 = 36

4x+4-4 = 36-4 ... subtract 4 from both sides

4x = 32

4x/4 = 32/4 .... divide both sides by 4

x = 8

With x = 8, we can say

2x+1 = 2*8+1 = 17 is the first odd integer

2x+3 = 2*8+3 = 19 is the next odd integer

17+19 = 36, so the answer is confirmed

--------------------

A non-algebraic way to get the answer is to divide 36 in half to get 18

Then try odd integers around 18, so 19+21 = 40 but thats too high

then maybe try 15+17 = 32, thats too low

17+19 = 36 is right on the money

DanielleElmas [232]2 years ago
4 0

how many consecutive odd integers are there, or else I can't help you

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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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