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:
51,111%
Step-by-step explanation:
We know that 45% have a dog but no cat. While 23% have both a dog and a cat.
Then Pr(X has cat|X has dog) = Pr(X has dog and cat)/Pr(X has dog)
= Pr(X has dog and cat)/Pr(X has dog) = 0.23/0.45 = 0.51 ≈ 51,111%
The probability that a student’s family owns a cat if the family owns a dog is 51,111%
Answer: 75
work: n/5 + -6 = 9
n/5 = 9+ 6
n/5 = 15
n = 15 x 5
n= 75
1) 4
2) 16
3) 16
4) 6
5)12
6) 4
