5000(1+.085/4)^n=1000
to find n use log:
1.02125^n
N X log1.02125
when computed=32.9
round to 33 quarters compounded
33/4
equals to 8 years 3 months
Y = a (b^x)
1) x = 0, y = 3500
=> 3500 = a (b^0)
=> 3500 = a (1)
=> a = 3500
2) growth percent = 3.5% => growth rate = 1.035 = b
=> y = 3500 (1.035)^ x
3) year 2020 => x = 2020 - 2000 = 20
4) y = 3500 (1.035)^ 20 = 6964
Answer: 6964.
3x - 18 = 42
3x = 42 + 18
3x = 60
x = 60/3
x = 20
Answer:
$0.025x² . . . where x is a number of percentage points
Step-by-step explanation:
The multiplier for semi-annual compounding will be ...
(1 + x/2)² = 1 + x + x²/4
The multiplier for annual compounding will be ...
1 + x
The multiplier for semiannual compounding is greater by ...
(1 + x + x²/4) - (1 + x) = x²/4
Maria's interest will be greater by $1000×(x²/4) = $250x², where x is a decimal fraction.
If x is a percent value, as in x = 6 when x percent = 6%, then the difference amount is ...
$250·(x/100)² = $0.025x² . . . where x is a number of percentage points
_____
<u>Example</u>:
For x percent = 6%, the difference in interest earned on $1000 for one year is $0.025×6² = $0.90.
Answer:
16
Step-by-step explanation:
2/0.25 = 8
2 x 8 = 16