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MA_775_DIABLO [31]
2 years ago
10

John invested $45 000 in the investment account that pays an interest rate of 17.75%p.a. compounding bi-annually for 10 years. A

fter five and half years from the investment starting date, the interest rate changed from 17.75% to 18.25% p.a. compounding quarterly apply the formula, E=A (1+i/100)^n. what will be the compound amount will John have after 10 years
Mathematics
1 answer:
Softa [21]2 years ago
4 0
I will rewrite the formula on my own way.
The formula is
A=p (1+r/k)^kt
A future value?
P present value 45000
R interest rate 0.1775
K bi-annually 2
T time 5.5 years for the first period
A=45,000×(1+0.1775÷2)^(2×5.5)
A=114,662.75225375

Now use the formula again where the present value is 114,662.75225375 and the interest rate is 18.25% compounded quarterly (4).
The time of the rest of the period is 4.5 years (10-5.5)

A=114,662.75225375×(1+0.1825÷4)^(4×4.5)
A=255,970.32....answer
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the 41st term of the arithmetic sequence is -539 and the 54th term is -708. Find the value of the 80th term
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Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m
suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week                              1           2            3           4            5

             Water Level (inches) − 1 1 4     1 3 8      2 1 2      -1 5 8      -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = \frac{138 - 114}{2 - 1} = \frac{24}{1} = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = \frac{212 - 114}{3 - 1} = \frac{98}{2} = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = \frac{-158 - 114}{4 - 1} = -\frac{272}{3} = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = \frac{-134 - 114}{5 - 1} = -\frac{248}{4} = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = \frac{212 - 138}{3 - 2} = \frac{74}{1} = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = \frac{-158 - 138}{4 - 2} = -\frac{296}{2} = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = \frac{-134 - 138}{5 - 2} = -\frac{272}{3} = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = \frac{-158 - 212}{4 - 3} = -\frac{370}{1} = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = \frac{-134 - 212}{5 - 3} = -\frac{346}{2} = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = \frac{-134 + 158}{5 - 4} = \frac{24}{1} = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0
2 years ago
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