10m + n^2/4
m = 5
n = 4
We can now substitute numbers in for the variables.
10(5) + (4)^2/4
50 + 4^2/4
50 + (4 * 4)/4
50 + 16/4
50 + 4
= 54
Answer:
11 cats
Step-by-step explanation:
Given :
Jane's initial = 23
My initial = 2
Jane gives out 5 cats to me ;
Jane's new = 23 - 5 = 18
My new = 2 + 5 = 7
Number of cats Jane has than me :
Jane's new - My new
18 - 7 = 11 cats
Second question is incomplete. We aren't told what to calculate.
Simplified would be 2a+2b
Since the standard deviation is not given, we assume the process is a Poisson process, with mean 12000 miles.
From Poisson distribution tables, we see that with a value of lambda(mean)=12000, the lower tail probability for x=12140 miles is 0.9000105, which means that the 90 percentile is 12140 miles.
The lower-tail probability of 9000 miles is 6.2*10^(-181) which is essentially zero.
NOTE: if the problem were modelled with a normal distribution, the probabilities would be completely different.
Answer:
24
Step-by-step explanation:
Divide 40 by 10 and multiply your answer by 6