Answer is C
hope this helps!
Answer:
$138,345
Step-by-step explanation:
This is a compound decline problem, which will be solve by the compound formula:

Where
F is the future value (value of house at 2030, 14 years from 2016)
P is the present value ($245,000)
r is the rate of decline, in decimal (r = 4% = 4/100 = 0.04)
t is the time in years (2016 to 2030 is 14 years, so t = 14)
We substitute the known values and find F:

Rounding it up, it will be worth around $138,345 at 2030
Answer:
a) $86.11
b) $467.33
Step-by-step explanation:
no. of gallons saved 'n' per year
n = (10000/33) - (10000/36)
n = 2500/99
Amount saved = 3.41×n = 3.41×2500/99
= $86.11
b) over 5 years:
86.11( 1 + (1.041) + (1.041)² + (1.041)³ + (1.041)⁴ )
= 86.11(5.427157431)
= $467.33
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