Answer:
f(5)= 78.5. I hope this helps!!
Answer:
-6.134 to +6.134
Step-by-step explanation:
given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution
X is Normal with mean =6.1 and std dev = 1 unit
We are to determine the range of values containing 70% of the population of x
We know that normal distribution curve is bell shaped symmetrical about the mean.
So to find 70% range we can use 35% on either side of the mean
Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034
Hence corresponding x value is

i.e. 70% values lie between
-6.134 to +6.134
The formula of the future value of annuity due is
Fv=pmt [(1+r/k)^(kn)-1)÷(r/k)]×(1+r/k)
Fv future value 1250675
PMT semiannual payment?
R interest rate 0.045
K compounded semiannual 2
N time 30 years
Solve the formula for PMT
PMT=Fv÷ [(1+r/k)^(kn)-1)÷(r/k)]×(1+r/k) Plug in the formula
PMT=1,250,675÷((((1+0.045
÷2)^(2×30)−1)÷(0.045÷2))×(1+0.045÷2))
=9,828.44...Answer
Hope it helps!