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CaHeK987 [17]
3 years ago
10

The domain for f(x) and g(x) is the set of all real numbers. Let f(x) = 3x + 5 and g(x) = x2. Find g(x) − f(x).

Mathematics
1 answer:
kifflom [539]3 years ago
8 0

Answer:

3x^2 + 8  

Step-by-step explanation:

Given functions,

f(x) = 3x + 8 ----(1)

g(x) = x^2 -----(2)

Since,

(fog)(x) =  f( g(x) )      ( Composition of functions )

=f(x^2)             ( From equation (2) )

=3x^2 + 8          ( From equation (1) )

Hence,

(fog)(x)=3x^2 + 8

Step-by-step explanation:

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Answer:

a) We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

b) We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

Step-by-step explanation:

Part a

We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

Part b

We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

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Step-by-step explanation:

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Answer:

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