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ozzi
3 years ago
15

Which expression is equivalent to -6-(-2)?

Mathematics
2 answers:
Andre45 [30]3 years ago
8 0

Answer:

-2+6

Step-by-step explanation:

-6-(-2)

-6+2

-4

-2+6

=4

const2013 [10]3 years ago
5 0
The answer is -4 because two negatives is the same as adding so -6 + 2 = -4
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Yesterday Howard made $7 an hour while working. He also got a $10 tip. Let h
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m = 7h + 10

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Since m is representing the total amount of money he earned, we will leave that on one side. If h is the number of hours we worked, the coefficient of h is 7 because he earns 7 dollars an hour. The ten dollar tip is independent from the variables.

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The Jones family has 2 kids, 1 boy and 1 girl. Suppose that for each birth, the probability of a boy birth is
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A bag contains 5 red marbles 8 green marbles 6 orange marbles and 11 yellow marbles what is the probability of randomly choosing
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3 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

5 0
3 years ago
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