60
30 *2
15 * 2 * 2
5 * 3 * 2 * 2
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
It is £15 for the words, but I don’t understand the second part
Answer:
37 dimes and 10 nickels
Step-by-step explanation:
let d = # dimes
let n = # nickels
we can set up a system of equations:
n + d = 47
.05n + .10d = 4.2
if we solve the first equation for 'n' we get:
n = 47-d
now we can substitute this in for 'n' in the second equation:
.05(47-d) + .10d = 4.2
2.35 - .05d + .10d = 4.2
2.35 + .05d = 4.2
subtract 2.35 from each side to get:
.05d = 1.85
d = 1.85÷.05
d = 37
if d+n = 47 and d=37 then n = 10
Check:
.05(10) + .1(37) should equal 4.2
.50 + 3.7 = 4.2 [It Checks Out]
Answer:
4 
Step-by-step explanation:
so we see a wholoe number "4", lets set that aside
we are left with 0.75 right
we can also write this as
75/100
we can simplify this by dividing 25 on both sides so it will becocme:
3/4
*Remember we have to bring the "4" we set aside earlier so final answer is*
4 and 3/4
