9514 1404 393
Answer:
558
Step-by-step explanation:
Let p represent the student population at Edison Jr High. Then we know ...
(4/9)p = 248 . . . . . 248 students is 4/9 of the student population
p = 248(9/4) = 558 . . . multiply the equation by 9/4
558 students attend Edison Jr. High.
I would go for Obtuse from the 120° angle, Acute from the either the 20° and 40°, and scalene since all angles are unequal
Answer:
the answer is 1760
Step-by-step explanation:
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.
Use the expression for y in the second equation.
.. 3x +4(3 -(1/2)x) = 1
.. 3x +12 -2x = 1
.. x +12 = 1
.. x = -11
