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If f(x) = x2 + 1 and g(x) = x – 4, which value is equivalent to mc024-1.jpg?

Harrizon [31]

The complete question in the attached figure

we have that
f(x) = x²<span> + 1
g(x) = x – 4

step 1
find </span>(f o g)(x)
(f o g)(x)= (x - 4)² + 1(f o g)(x) = x² - 8x + 16 + 1
(f o g)(x) = x² - 8x + 17

step 2
find (f o g)(10)
(f o g)(10) = 10² - 8*(10) + 17
(f o g)(10 = 100 - 80 + 17
(f o g)(10)= 37

the answer is 37

4 0

3 years ago

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RSTU is a rectangle with vertices at R(-3,1), S(1 ,1), T(-3,3) and U (1, 3). Describe two lines of reflection that would carry t

miss Akunina [59]

Answer:

The two lines of reflection are $x=-1$ and $y=2$

Step-by-step explanation:

I will attach a file with the graph of the rectangle and the lines of reflection.

We know that RSTU is a rectangle with vertices at $R=(-3,1)$ , $S=(1,1)$ ,

$T=(-3,3)$ and $U=(1,3)$

The first step is to draw the vertices on the plane and them the rectangle.

The rectangle is a symmetrical figure. Therefore if we want to carry the rectangle onto itself using a line of reflection this line must goes through the centroid of the rectangle.

Given that we have a rectangle whose width is $a$ and its length is $b$ its centroid will be place at $(\frac{a}{2},\frac{b}{2})$ .

Looking at the graph and using this information we find that the two lines of reflection are $x=-1$ and $y=2$ .

6 0

3 years ago

7. Cody bought 4 games for $35.90. Which equation can be used to determine

Jlenok [28]

Answer:

35.90÷4 is the equation for one video game

4 0

3 years ago

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Tina placed a 12 meter rope along one side of the bicycle path. She hung a ribbon on each end of the rope and every 3 meters in-

olga_2 [115]

Answer:

4

Step-by-step explanation:

If the tape is 12m,

she hung the tape in between every 3m, and at both ends

0, 3, 6, 9, 12

these are the lengths she hung ribbons at

Total number of ribbons = 4

A is correct

5 0

3 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

koban [17]

Answer:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) P=0.42

c) P=0.08

d)

X | P(X)

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

0 | 0.42

1 | 0.50

2 | 0.08

e) E(x)=0.66

s.d.=0.62

Step-by-step explanation:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) The probability of losing the first game is

$P(G_1=L)=1-P(G_1=W)=1-0.4=0.6$

The probability of losing the second game, given that the first game was lost is:

$P(G_2=L|G_1=L)=1-P(G_2=W|G_1=L)=1-0.3=0.7$

So the probability of losing both games is:

$P(G_2=L\&G_1=L)=P(G_1=L)*P(G_2=L|G_1=L)=0.6*0.7=0.42$

c) The probability of winning both games is:

$P(G_2=W\&G_1=W)=P(G_1=W)*P(G_2=W|G_1=W)=0.4*0.2=0.08$

d) The variable X can take values 0, 1 and 2.

X=0 is when both games are lost. This happens with probability P=0.42.

X=2 is when both games are won. This happens with probability P=0.08.

X=1 is when one game is won and the other is lost. This happens with probability P=1-0.42-0.08=0.50.

Then the table of probabilities become:

X | P(X)

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

0 | 0.42

1 | 0.50

2 | 0.08

e) The expected value is:

$E(x)=\sum p_ix_i=0.42*0+0.50*1+0.08*2=0.00+0.50+0.16=0.66$

The variance and standard deviation of x are:

$V(x)=\sum p_i(x_i-E(x))^2\\\\V(x)=0.42(0-0.66)^2+0.50(1-0.66)^2+0.08(2-0.66)^2\\\\V(x)=0.42*0.4356+0.50*0.1156+0.08*1.7956\\\\V(x)=0.183+0.058+0.144=0.385\\\\\\\sigma=\sqrt{V(x)}=\sqrt{0.385}=0.62$

The standard deviation can

5 0

3 years ago