Answer:lol so true
Step-by-step explanation:
25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Use the formula of the present value of an annuity ordinary which is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value 5500
PMT monthly payment?
R interest rate 0.115
K compounded monthly 12
N time 5years
Solve the formula for PMT
PMT=Pv÷ [(1-(1+r/k)^(-kn))÷(r/k)]
PMT=5,500÷((1−(1+0.115÷12)^(
−12×5))÷(0.115÷12))
=120.95
So the answer is C
Hope it helps!
Answer:
35%
Step-by-step explanation:
Your answer would look like this