First we will find the interest on:
P = $235 principal
t = 2 years
r = 0.1415 annual rate
A = future value
I = A - P the interest
A = P(1 + r)^t
A = 235(1 + 0.1415)^2
A = $306.21
I = A - P
I = $306.21 - $235
I = $71.21
the interest was $71.21.
Next lets find the lifetime cost value:
Lifetime cost value = 306.21 + 5*1.56*52 + 5*0.78*52 = $914.61 (considering that 1 year = 52 weeks)
Now lets find the percentage what percentage the interest is of the lifetime cost:
(71.21/914.61)*100 = 7.79%
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N/d = 2/3 -------------------> 3n = 2d
(n + 2)/(d + 2) = 3/4 ------> 4n + 8 = 3d + 6
Answer:
so principal amount is $7862.07
Step-by-step explanation:
Given data
rate = 7*1/4 % = 29/4 %
interest = $190
time = 4 months = 4/12 year
to find out
principal
solution
we know the simple interest formula i.e.
interest = ( principal × rate × time ) /100 ..................1
now put all value rate time and interest in equation 1 we get interest here
interest = ( principal × rate × time ) /100
190 = ( principal × 29/4 × 4/12 ) /100
principal = 190 × 12 ×100 / 29
principal = 7862.068966
so principal amount is $7862.07
Answer:
n2^3 bacteria, n being the initial population.
Step-by-step explanation:
compound growth equation: P=n2^t
P is total population
n is initial population
t is time
since it doubles every second hour, and there are 6 hours
6hours/2hours= doubles 3 times, thats why t=3.
