6j + 1/3n = 134...multiply by 3....18j + n = 402
1/3j + n = 31....multiply by 3....j + 3n = 93
the multiplying by 3 is optional...I just did it because it is easier to work with equations when there are no fractions.
18j + n = 402....multiply by -3
j + 3n = 93
----------------
-54j - 3n = - 1206 (result of multiplying by -3)
j + 3n = 93
----------------add
-53j = - 1113
j = -1113/-53
j = 21
j + 3n = 93
21 + 3n = 93
3n = 93 - 21
3n = 72
n = 72/3
n = 24
so Jasons collection (j) consists of 21 books and Nathans collection (n) consists of 24 books
2, 3, 5. I just solved them and they look equivalent.
Step-by-step explanation:
I'm not sure if this will help but to find out how much coffee 4 people would need you would have to divide 4 into 5 which is 1 1/5
EQUATION-- 4/5= X
I chose that because / means divide 5 into 4 would give you x we don't know what x is until we actually do the problem
Even if im not correct I hope it helped a bit
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
