It could be 443, try simplifying to the nearest hundreaths, hope this helps!
Answer:
Trent has 6 quarters and 5 dimes.
Step-by-step explanation:
Consider the provided information.
Let q = Number of quarters
d = Number of dimes.
Trent has 11 quarters and dimes.
The worth of a quarter is 25 cents.
The worth of a dimes is 10 cents
Substitute the value of q in above equation.
Substitute the value of d in
Hence, Trent has 6 quarters and 5 dimes.
<span>$84 to 12%
</span>10.08
is the answer
96: 1,2,3,4,6,8,12,16,24,32,48,96
18: 1,2,3,6,9,18
GCF: 6
But, if you're asking what 96 + 18 it's 114
Answer:
C
Step-by-step explanation:
This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.
Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:
10 C 6 = 10!(6!4!)
If we operate, knowing that for any number x, x!=x*(x-1)*(x-2)*...*1
10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]
10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]
We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3
10 C 6 = 10*9*8*7*/ [(4*3*2*1)] = 10*3*3*8*7*/ [(8*3*1)]
We can eliminate the 8s and one of the 3s on the numerator with the one on the denominator:
10 C 6 = 10*3*7*/1 = 210/1= 210
So, option C
