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3 years ago

Two functions are shown in the table below.

Mathematics

2 answers:

pashok25 [27]3 years ago

6 0

Answer:

x=4

Step-by-step explanation:

umka21 [38]3 years ago

5 0

Given:
f(x) = - x² + 4x + 12
g(x) = x + 8

f(1) = -(1)² + 4(1) + 12 = -1 + 4 + 12 = -1 + 16 = 15
f(2) = -(2)² + 4(2) + 12 = -4 + 8 + 12 = -4 + 20 = 16
f(3) = -(3)² + 4(3) + 12 = -9 + 12 + 12 = - 9 + 24 = 15
f(4) = -(4)² + 4(4) + 12 = -16 + 16 + 12 = -16 + 28 = 12
f(5) = -(5)² + 4(5) + 12 = -25 + 20 + 12 = -25 + 32 = 7
f(6) = -(6)² + 4(6) + 12 = -36 + 24 + 12 = -36 + 36 = 0

g(1) = 1 + 8 = 9
g(2) = 2 + 8 = 10
g(3) = 3 + 8 = 11
g(4) = 4 + 8 = 12
g(5) = 5 + 8 = 13
g(6) = 6 + 8 = 14

f(x) = g(x) when x = 4
f(4) = g(4)
12 = 12

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Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g

Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

$E[g(x)] = \sum g(x)p(x)$

So, $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

$g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also, $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

$g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

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