1. 5x - 7 = 48
5x = 48 + 7 = 55
x = 55/5 = 11
2. 7n - 2 = 54
7n = 54 + 2 = 56
n = 56 / 7 = 8
3. -4y + 7 = -21
-4y = -21 - 7 = -28
y = -28 / -4 = 7
4. x/2 + 5 = 12
x/2 = 12 - 5 = 7
x = 7 * 2 = 14
5. x/3 - 7 = 13
x/3 = 13 + 7 = 20
x = 20 * 3 = 60
6. -7 - 2 = 33 <- did you type this wrong?
-7x - 2 = 33
-7x = 33 + 2
-7x = 35
x = 5
7. -3x - 8 = -41
-3x = -41 + 8 = -33
x = -33 / -3 = 11
8. 3n - 7 = 35
3n = 35 + 7 = 42
n = 42 / 3 = 14
9. 7y + 9 = -47
7y = -47 - 9 = -56
y = -56 / 7 = -8
10. -8x + 4 = 68
-8x = 68 - 4 = 64
x = 64 / -8 = -8
11. 4n - 7 = -43
4n = -43 + 7 = -36
n = -36 / 4 = -9
12. 5x - 7 = 108
5x = 108 + 7 = 115
x = 115 / 5 = 23
13. 8 - 5w = -37
-5w = -37 - 8 = -45
w = -45 / -5 = 9
14. 42 = 18 + 4n
4n = 42 - 18 = 24
n = 24 / 4 = 6
15. 70 - 6x = 40
-6x = 40 - 70 = -30
x = -30 / -6 = 5
Answer:
6 of each
Step-by-step explanation:
1.50 x 6= 9
1 x 6= 6
9+6=15
It would be: 6/25 * 100 = 6 * 4 = 24%
In short, Your Answer would be Option C
Hope this helps!
Step-by-step explanation:
The data below is what was provided in the question and it is what I solved the question with
P(A1) = 0.23
P(A2) = 0.25
P(A3) = 0.29
P(A1 n A2 ) = 0.09
P(A1 n A3) = 0.11
P(A2 n A3) = 0.07
P(A1 n A2 n A3) = 0.02
a
P(A2|A1) = P(A1 n A2)/P(A1)
= 0.09/0.23
= 0.3913
We have 39.13% confidence that event A2 will occur given that event A1 already occured
b.)
P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)
= 0.02/0.23
= 0.08695
We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.
C.
P(A2 u A3|A1)
= P(A1 n A2)u(A1 n A3)/P(A1)
= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)
= (0.09+0.11-0.02)/0.23
= 0.18/0.23
= 0.7826
We have 78.26% chance of A2 or A3 happening given that A1 has already occured.
