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Ratling [72]
2 years ago
6

Please can someone help with rearranging formulae? Will mark brainliest!

Mathematics
1 answer:
3241004551 [841]2 years ago
6 0

Answer:

look at the picture ..........

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{7,14}

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Domain is always the X= Values

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3 years ago
Let f(x)=100−10+e−0.1x .
lianna [129]
What are you solving for?
If you are solving for × then I hope this helps

Let's solve for x.

f(x)= 100−10+e−0.1x

Step 1: Add 0.1x to both sides.

xf+0.1x=−0.1x+e+90+0.1x

xf+0.1x = e+90

Step 2: Factor out variable x.

x(f+0.1) = e+90

Step 3: Divide both sides by f+0.1.

x(f+0.1)/ f+0.1 =e+90/ f+0.1

x=e+90/ f+0.1

Answer:

x= e+90/ f+0.1
4 0
4 years ago
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How to write equation
DaniilM [7]

Answer:

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6 0
3 years ago
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

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

∴ $g(X) = xe^{tx}$

7 0
3 years ago
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