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Ludmilka [50]
3 years ago
9

Abdul ordered some books online and spent a total of $133. Each book cost $8 and he paid a total of $5 for shipping. How many bo

oks did he buy?
Mathematics
2 answers:
Vsevolod [243]3 years ago
8 0

Answer:

8x+5=133

8x=128

x=16

Step-by-step explanation:

16 books

Ivahew [28]3 years ago
6 0

Answer:

He bought 16 books

Step-by-step explanation:

cost = flat fee + books * cost per book

133 = 5 +b*8 where b is the number of books

133 = 5+8b

Subtract 5 from each side

128 = 8b

Divide each side by 8

128/8 = 8b/8

16 = b

He bought 16 books

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Which point is Not on a circle with a center of (0,0) and a radius of 10?
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It is A because the radius is 10 which means that if x=0 then y must equal 10. (0,5) would be inside of the circle. If this doesn't make sense graph the circle and the points and you'll see what I mean.
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An element with mass 780 grams decays by 17.1% per minute. How much of the element is remaining after 19 minutes to the nearest
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Step 1
Formulate a recursive sequence modeling the number of grams after n  minutes.
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3 years ago
Write and solve the equation. Four times the sum of x and 5 equals twice the difference of 10 and that same number.
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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

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