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riadik2000 [5.3K]
3 years ago
13

What is the image of (9,0) after a dilation by a scale factor of 2 centered at the origin?

Mathematics
1 answer:
Advocard [28]3 years ago
6 0
(5,7). Because the origin centers 2 by dilation.
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ohaa [14]
To do this, you can just multiply across. 3 times 1 equals 3 so your numerator is 3. 2 times 4 equals 8. Then, your answer would be 3/8. Hope this helps :)
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Jonathan sold five times as many computers as Joe sold last year. In total,
Rom4ik [11]

Answer:

Step-by-step explanation:

Let the number of computers sold by Joe be represented by variable X

If Jonathan sold 5 times as many computers , it means he sold 5 times X= 5X

Total number of computers sold by both Jonathan and Joe = 78

Hence, X+5X =78

= 6X = 78

Divide both sides by 6 to solve for X;

X = 78/6

X = 13

And 5X = (5* 13) = 65

Therefore, Jonathan sold 65 computers while Joe sold 13

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3 years ago
Lewis needs 1/8 quart of paint to cover 10 square yards. How many square yards can he paint with 4 quarts of paint?
nalin [4]

Answer:

Lewis can paint 320 square yards.

Step-by-step explanation:

1 quart covers 80 square yards

8 x 10=80

4 quarts covers 320 square yards

80 x 4=320

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2 years ago
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Crazy boy [7]

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2 years ago
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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
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