The following federal tax table is for biweekly earnings of a single person. A 9-column table with 7 rows is shown. Column 1 is labeled If the wages are at least with entries 720, 740, 760, 780, 800, 820, 840. Column 2 is labeled But less than with entries 740, 760, 780, 800, 820, 840, 860. Column 3 is labeled And the number of withholding allowances is 0, the amount of income tax withheld is, with entries 80, 83, 86, 89, 92, 95, 98. Column 4 is labeled And the number of withholding allowances is 1, the amount of income tax withheld is, with entries 62, 65, 68, 71, 74, 77, 80. Column 5 is labeled And the number of withholding allowances is 2, the amount of income tax withheld is, with entries 44, 47, 50, 53, 56, 59, 62. Column 6 is labeled And the number of withholding allowances is 3, the amount of income tax withheld is, with entries 26, 28, 31, 34, 37, 40, 43. Column 7 is labeled And the number of withholding allowances is 4, the amount of income tax withheld is, with entries 14, 16, 18, 20, 22, 24, 26. Column 8 is labeled And the number of withholding allowances is 5, the amount of income tax withheld is, with entries 1, 3, 5, 7, 9, 11, 13. Column 9 is labeled And the number of withholding allowances is 6, the amount of income tax withheld is, with entries 0, 0, 0, 0, 0, 0, 1. A single person earns a gross biweekly salary of $840 and claims 5 exemptions. How will federal taxes affect his net pay? a. Federal taxes will not affect the net pay. b. Net pay will increase $11 from the gross pay. c. Net pay will decrease $11 from the gross pay. d. Net pay will decrease $13 from the gross pay. Please select the best answer from the choices provided A B C D Mark this and retur

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

nope lol

step-by-step explanation:

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step-by-step explanation: