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Vitek1552 [10]
3 years ago
12

Somebody please help me with this !!!

Mathematics
1 answer:
Damm [24]3 years ago
7 0

Step-by-step explanation:

Case 1 :

3x + 2y = 6 \\ 4x + y = 2 \\ 3x = 6 - 2y \\ x =  \frac{6 - 2y}{3}  = \\ x = 2 -  \frac{2y}{3}  \\ 4(2 -  \frac{2y}{3} ) + y = 2 \\ 8 -  \frac{8y}{3}  + y = 2 \\  - 1.67y =  - 6 \\

y = 3.59 \\ x = 2 -  \frac{2y}{3}  \\x =  2 -  \frac{2 \times 3.59}{3}  \\  \\ x =  - 0.39

Case 2 :

3x + 2y = 6 \\ 4x  -  y = 2 \\ x = 2 -  \frac{2y}{3}  \\ 4(2 -  \frac{2y}{3} ) - y = 2 \\ 8 -  \frac{8y}{3}  - y = 2 \\   - 3.67y=  - 6 \\ y = 1.63

x = 2 -  \frac{2y}{3}  \\ x = 2 -  \frac{2 \times 1.63}{3}  \\ x = 2 - 1.087 \\ x = 0.913

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There are 16 girls I a school club. The number of girls is 4 more than twice the number of boys. Find the number of boys.
olga_2 [115]

6 boys

16-4=12

12/2=6

It's not that hard of a problem you just have to read in-between the lines

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3 years ago
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You roll a pair of honest dice. If you roll a total of 7, you win $18; if you roll a total of 11, you win $54; if you roll any o
amm1812
Answer: -1
The negative value indicates a loss

============================================================

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

-----------------------------

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.


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