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tensa zangetsu [6.8K]
3 years ago
7

A box contains five slips of paper. Each slip has one of the number 4, 6, 7, 8, or 9 written on it and all numbers are used. The

first player reaches into the box and draws two slips and adds the two numbers. If the sum is even, the player wins. If the sum is odd, the player loses.
a. What is the probability that the player wins?

b. Does the probability change if the two numbers are multiplied? Explain.
Mathematics
1 answer:
max2010maxim [7]3 years ago
6 0
A. If the player chooses 6, 4, and 8 each time, they have 3/5 chances to win.

6+4= 10

6+8= 14

8+4= 12

These are all even so as long as they get one of these combinations they will win

b. All three combinations still work when multiplied

6*4=24
6*8=48
8*4= 32
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Kindly find complete question attached below

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3 years ago
PLEASE HELP 2O POINTS
vovangra [49]
C. The first is a solution, but the second is not.

5x - y/3 = 13 ; (2,-9)
substitute the letters:
5(2) - (-9/3) = 13 
10 - (-3) = 13 : note that deducting a number with a negative sign turns both sign as positive.
10 + 3 = 13 ;
13 = 13

5x - y/3 = 13 ; (3,-6)
5(3) - (-6/3) = 13
15 - (-2) = 13
15 + 2 = 13
17 = 13 not equal. not a solution

hope i could help
4 0
3 years ago
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