Costs associated with the manufacture of miniature high-sensitivity piezoresistive pressure transducers is $82,000 per year. A clever industrial engineer found that by spending $18,000 now to reconfigure the production line and reprogram two of the robotic arms, the cost will go down to $55,000 next year and $57,000 in years 2 through 5. Using an interest rate of 9% per year, determine the present worth of the savings due to the reconfiguration. The present worth of the savings is determined to be $

PW = $62750.075

Hence, the present worth of the savings is $62750.075

Explanation:

cost of manufacturing is = $82000

Spending now amount = $18000

Cost will go down to = $55,000 in year 1

Cost will go down to = $57,000 from year 2 to year 5

By investing $18000 now,

amount saved in year 1= ($82000 - $55000) = $27000

Amount saved in year 2 to year 5 =  ($82000 - $57000) = $25000

Time period = 5 years

Rate of interest = 9% per year.

We can use the following formula to calculate the present worth of the savings:

PW = -P + F (P/F, i, n)  + [A(P/A , i, n)] (P/F,i,n)

Where,

P = Initial Amount Spent = $18000

F = Amount Saved in Year 1 =  $27000

A = Amount Saved in Year 2 to 5 = $27000

i = Rate of interest = 9%

n = Time period = 5 years

Just plug in these values into the formula to get the results.

PW = -$18000 + $27000(P/F, 9%, 5) + [$25000(P/A , 9%, 5)] (P/F,9%,5)

Where,

(P/F, 9%, 5) = 0.6499  

(P/A , 9%, 5) = 3.890

Use interest tables.

Just Plugging in these values in to the formula.

PW = -$18000 + $27000 x (0.6499)  + [$25000 x (3.890) ] x (0.6499)

PW = -$18000 + 17547.3 + [97250] x (0.6499)

PW = -$18000 + 17547.3 + 63202.775

PW = $62750.075

Hence, the present worth of the savings is $62750.075

1: b     the use of non-standard grammar

2: c     strong and willful.

its correct i took the test have a nice day: )

you have to put the parentheses like this:

in fact, you have to perform the operation inside parentheses first, and then proceed in order: