If you finish the question I may be able to answer it. ^^
Using a binomial distribution considering there's a 30% chance it will rain on any of the three days:
<span>The probability of it raining on 0 days is (1)(0.7)(0.7)(0.7) = 34.3%. </span>
<span>The probability of it raining on 1 day is (3)(0.3)(0.7)(0.7) = 44.1%. </span>
<span>The probability of it raining on 2 days is (3)(0.3)(0.3)(0.7) = 18.9%. </span>
<span>The probability of it raining on 3 days is (1)(0.3)(0.3)(0.3) = 2.7%. </span>
<span>There's a 65.7% chance that it will rain at least once over the three-day period.</span>
Answer:
(a)9×3=37' 97 and 8 are factors of 37
Current amount in account
P=36948.61
Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n
Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11
We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000
We can solve this by trial and error.
The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000. It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.
So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.
So it takes 15.00 years to reach the goal of 280000 years.
