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butalik [34]
3 years ago
5

The formula S = A((1+r)t+1−1r) S = A 1 + r t + 1 - 1 r models the value of a retirement account, where A = the number of dollars

added to the retirement account each year, r = the annual interest rate, and S = the value of the retirement account after t years. If the interest rate is 11%, how much will the account be worth after 15 years if $2200 is added each year? Round to the nearest whole number.
Mathematics
2 answers:
Roman55 [17]3 years ago
8 0

Answer:

$86217.8866411

Step-by-step explanation:

Formula : S=A\frac{(1+r)^{(1+t)}-1}{r}

Where

A = the number of dollars added to the retirement account each year

r = the annual interest rate

S = the value of the retirement account after t years.

Given :

interest rate r= 11% =0.11

Amount (A) = $2200

Time(t) = 15 years

Substituting values in the formula :

S=2200\frac{(1+0.11)^{(1+15)}-1}{0.11}

S=86217.8866411

Thus the account will be worth $86217.8866411 after 15 years if $2200 is added each year.

tatiyna3 years ago
4 0
So this is how you will arrive to the answer:

The following formula models the value of a retirement account,

S = (A [ ( 1 + r ) ^ (t + 1) - 1] / r)

wherein:

A = number of dollars added to the retirement account (each year)

r = annual interest rate

s = value of the retirement account after t years

The question is:

If the interest rate is 11% then how much will the account be worth after 15 years if $2200 is added each year?

Round to the nearest whole number.

Solution:

The said formula contains the term t + 1 instead of the usual "t". Means that the formula applies only in the situation where the money is invested at the beginning of the year instead of the usual practice at the end

Given:

A = 2200
r = 0.11
t = 15

The accumulated amount:
F = A ((1 + r) ^ (t+1) - 1 / r

Substitute:

F = 2200 (1.11 ^ (15 + 1 ) - 1) /0.11
F = 86217.88664 

If money is invested at the end of the year, then F = 80476.49, the difference being the investment of an extra 2200 over 15 years.
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Advocard [28]

Answer:

A). 15

B). -1.2

Step-by-step explanation:

A). [10(x+3)]-20 = 160

10x+30-20 = 160

10x+10 = 160

10x = 160-10

10x = 150

x = 150/10

x = 15

B). 5{[4(8+(5x))]-9} = 85

5(32+20x+9) = 85

5(32+9+20x) = 85

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----------------------------------------------------------------------

HOPE THIS HELPS!!!

3 0
3 years ago
During his hike, Milt drank 1 liter of water and 1 liter of sports drink. How many milliliters of liquid did he drink?
S_A_V [24]

Answer: 2,000 milliliters.  

Step-by-step explanation:

Hi, to answer this question we have to add the amount of water and the amount of sports drink that he drank:

1 liter + 1 liter = 2 liters of liquid.

Finally, we have to convert the result into milliliters.

Since 1 liter = 1000 liters

We have to multiply the liters drank by 1000.

2 x 1000 = 2,000 milliliters.  

Feel free to ask for more if needed or if you did not understand something.  

5 0
3 years ago
Show all work to identify the asymptotes and zero of the function f of x equals 5 x over quantity x squared minus 25.
Lorico [155]

Answer:

  • asymptotes: x = -5, x = 5
  • zero: x = 0

Step-by-step explanation:

The function of interest is ...

  f(x)=\dfrac{5x}{x^2-25}=\dfrac{5x}{(x-5)(x+5)}

The asymptotes are found where the denominator is zero. It will be zero when either factor is zero, so at x = 5 and x = -5

__

The zeros are found where the numerator is zero. It will be zero for x = 0.

The asymptotes are x=-5, x=5; the zero is x=0.

4 0
3 years ago
What equation best models this data?(use y to represent the population of rabbits and t to represent the year, assuming that 201
liraira [26]

If we see the data closely, a pattern emerges. The pattern is that the ratio of the population of every consecutive year to the present year is 1.6

Let us check it using a couple of examples.

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Likewise, the rabbit population in the year 2011 is 80. The population increases to 128 the next year (2012). Again, \frac{128}{80}=1.6

We can verify the same ratio with all the data provided.

Thus, we know that the population in any given year is 1.6 times the population of the previous year. This is a classic case of a compounding problem. We know that the formula for compounding is as:

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Where F is the future value of the rabbit population in any given year

P is the rabbit population in the year "0" (that is the starting year 2010) and that is 50 in this question. (please note that there is just one starting year).

r is the ratio multiple with which the rabbit population increases each consecutive year.

n is the nth year from the start.

Let us take an example for the better understanding of the working of this formula.

Let us take the year 2014. This is the 4th year

So, the rabbit population in 2014 should be:

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This is exactly what we get from the table too.

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4 0
3 years ago
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zysi [14]

Answer:

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Step-by-step explanation:

the question is based on similarity and enlargement

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simplify to obtain2:5 as your linear scale factor...

if 5:150

2=?

2*150/5=60

7 0
3 years ago
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