Answer:
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Step-by-step explanation:
Let's define the events:
L: The student is proficient in reading
M: The student is proficient in math
The probabilities are given by:
![P (L) = 0.81\\P (M) = 0.74\\P (L\bigcap M) = 0.64](https://tex.z-dn.net/?f=P%20%28L%29%20%3D%200.81%5C%5CP%20%28M%29%20%3D%200.74%5C%5CP%20%28L%5Cbigcap%20M%29%20%3D%200.64)
![P (M\bigcap L^c) = P (M) - P (M\bigcap L) = 0.74 - 0.64 = 0.1\\P (M^c\bigcap L) = P (L) - P (M\bigcap L) = 0.81 - 0.64 = 0.17](https://tex.z-dn.net/?f=P%20%28M%5Cbigcap%20L%5Ec%29%20%3D%20P%20%28M%29%20-%20P%20%28M%5Cbigcap%20L%29%20%3D%200.74%20-%200.64%20%3D%200.1%5C%5CP%20%28M%5Ec%5Cbigcap%20L%29%20%3D%20P%20%28L%29%20-%20P%20%28M%5Cbigcap%20L%29%20%3D%200.81%20-%200.64%20%3D%200.17)
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Answer:
£25
Step-by-step explanation:
100 - 15 = 85
53.55 ÷ 0.85 = 63
63 - 38 = 25
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.
Answer:
x=![\frac{y2 }{9}](https://tex.z-dn.net/?f=%5Cfrac%7By2%20%7D%7B9%7D)
Step-by-step explanation:
2nd step: ![3\sqrt{x}=y](https://tex.z-dn.net/?f=3%5Csqrt%7Bx%7D%3Dy)
third step: ![\frac{3\sqrt{x} }{3} = \frac{y}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Csqrt%7Bx%7D%20%7D%7B3%7D%20%3D%20%5Cfrac%7By%7D%7B3%7D)
fourth step: ![\sqrt{x} =\frac{y}{3}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%3D%5Cfrac%7By%7D%7B3%7D)
answer: x=![\frac{y2 }{9}](https://tex.z-dn.net/?f=%5Cfrac%7By2%20%7D%7B9%7D)
Answer:
6.5
Step-by-step explanation:
first subtract 7 from both sides
2x=13
next divide both sides by 2
x=6.5