a1=25(0.3)^2, common ratio = 0.3
So sum of first n terms of a geometric series = a1(1-r^n)/(1-r)
So sum of frist 10 terms of the given series = 25(0.3)^2(1-0.3^10)/ (1-0.3)
So sum of n=2 to n=10 is : 25(0.3)^2(1=0.3^10)/(1-0.3) -25(0.3)^2= 0.96426673425
We are interested in the left-end area under the standard normal curve that has the area 0.025.
But the problem seems to be simpler than that:
If n = number of students in the class, and we know that 2.5% of these students failed the course, and that 5 failed, then we can surmise that
0.025n = 5. Solving for n by multiplying both sides by 40, we get n = 200.
There were 200 students in the class.
It is perfectly fair for a teacher to put more info into a problem statement than you really need to solve the problem. That seems to be what happened here.
You do 2x+9=3x-1 which is x=10 meaning the measurements for 1 and 3 are 29 degrees and the measurements for 2 and 4 and 151 degrees
Answer: 2,504?
Step-by-step explanation: baccically times the 2 only by 100, or at least that’s what I am seeing. LOL
Answer:
b.168
Step-by-step explanation:
