Answer: If the number of bills counted is odd, the ones digit is 5; if the number of bills is even, the ones digit is 0.
Explanation
1) Simulate the counting of the bills:
number of bills counted value ones digit
1 5 5
2 5 + 5 = 10 0
3 10 + 5 = 15 5
4 15 + 5 = 20 0
5 20 + 5 = 25 5
6 25 + 5 = 30 0
2) Pattern: you can see that the ones digit alreternate: 5, 0, 5, 0, 5, 0, ...
If the number of bills counted is odd the ones digit is 5, if the number of bills is even the ones digit is 0.
x =2 y =8
2x + 2y=20
2x=20-2y
x=10-y
Ahora len la otra ecuacion remplazamos x por (10-y)
-2(10-y)-6y=-52
-20+2y-6y=-52
2y-6y= -32 (-52+20= -32)
-4y=-32
-y=-8
y=8 x=10-y x=2
2*2+2*8=20
4+16=20
-2*2-6*8= -52
-4-48= -52
Well if their are only 2 section are able and they are all equal so it should be 2 parts
The answer is $4270
add 9500+1500+1870=12870 then subtract your savings 12870-6100=6770 then subtract your scholarship 6770-2500=4270
Answer: -1
The negative value indicates a loss
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Explanation:
Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)
There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die
There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18
Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event.
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9
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In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9
The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9
Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7
Add up those results
3+3+(-7) = -1
The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.
Note: because the expected value is not 0, this is not a fair game.
