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Anna35 [415]
3 years ago
14

A car travels 630 miles in 14 hours. At this rate, how far will it travel in 42 hours?

Mathematics
1 answer:
weqwewe [10]3 years ago
3 0

Assuming the car's speed \frac{630}{14}=45\mathrm{mph} does not change, the car will travel 45\cdot42=\boxed{1890} miles.

Hope this helps :)

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Which of the following number sentences is true?<br> A -9=9<br> B-9=-(9)<br> C 9=(-9)<br> D 9=-(-9)
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A and D are the true number sentences

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What is 4 divided by 3/4?
musickatia [10]

Answer:

16/3 or 5.33

Step-by-step explanation:

Cross multiply

4/1 * 3/4

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6 0
2 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

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Rise over run - 4/1=4
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