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AveGali [126]
3 years ago
10

The average age of engineering students at graduation is a little over 23 years. This means that the working career of most engi

neers is almost exactly 500 months. How much would an engineer need to save each month to accrue $5 million by the end of her working career? Assume a 9% interest rate, compounded monthly.
Engineering
1 answer:
RideAnS [48]3 years ago
6 0

Answer:

$916

Explanation:

To solve this, we use the formula

FV = P/i * [(1+i)^n - 1], where

FV = future value of the all the money invested, $5 million

n = time span, = 500 months

P = payment per month

I = interest rate, 9% by 12 months, = 0.0075

Considering that we have been given all in our question, then we substitute directly and solve. So we have,

5000000 = P/0.0075 * [(1+0.0075)^500 -1]

5000000 * 0.0075 = P * [1.0075^500 - 1]

37500 = P * [41.93 - 1]

37500 = P * 40.93

P = 37500/40.93

P = $916.20

Therefore, the engineer needs to save $916 in a month which is the accrued

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Explanation:

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3 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
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Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

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       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

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       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

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