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Fynjy0 [20]
3 years ago
13

Geometry lol i don’t know !!!

Mathematics
2 answers:
postnew [5]3 years ago
7 0
Bejeweled w whenever w wnekfkdkkwmwn ddndkkelenen
worty [1.4K]3 years ago
5 0

Answer: 1

Step-by-step explanation:

8x -1 = 4x + 3

-4x    - 4x

4x -1 = 3

     +1  +1

   4x=4

     x=1

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The graph of the function has slope of 1 and y-intercept of -6.​
True [87]

Answer:

= -5

Step-by-step explanation:

1-6=5

1-6

{Subtract the numbers:} 1-6=-5

= -5

6 0
2 years ago
What is the slope of the line that passes through the points (-3,12) and (6,24)?
Irina18 [472]

Answer: slope = 4/3

Step-by-step explanation:

(24-12)/(6- -3)

= 12/9

= 4/3

7 0
3 years ago
1/9,1/3,5/9 find the rule
anastassius [24]
It’s going up by 2/9 . if you did it for the first one it’d b 3/9 which is 1/3
3 0
3 years ago
How do you solve this ?
Alenkasestr [34]
See photo for solution

4 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
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