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kkurt [141]
3 years ago
14

(3,1)(2,0)(1,-1)(3,2)(2,1)(1,0) A: Function OB: Not a function​

Mathematics
1 answer:
levacccp [35]3 years ago
4 0

Answer:

Not a function

Step-by-step explanation:

Y values can repeat in a function but domain cannot. 3 is repeated twice, thus these values do not represent a function.

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Dmitrij [34]

2-1.75=.25

.25/2=.125/1

.125 is the answer

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How do you find the constant of variation when y=-(2/3) and x=3
Katyanochek1 [597]
You can't. If you think about the straight line on a graph, those numbers
describe a single point that the line goes through, and they don't tell you
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6 0
3 years ago
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9. Ellie is making a batch of pancakes
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Answer:

Step-by-step explanation:

Maybe your questions has some wrong numbers in it?  but, going by what's in your questions ,  just  1 cup of flour   :/   seems too easy

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3 years ago
Finding the geometric mean
grigory [225]

Answer:

To find the geometric mean of two numbers, just find the product of those numbers and take the square root!

Step-by-step explanation:

4 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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