Question

How much money should be invested now (rounded to the nearest cent), called the initial investment, in a Treasury Bond investment that yields 5.75% per year, compounded quarterly for 6 years, if you wish it to be worth $12,000 after 6 years?

Hint: A(t) = A0(1 + start fraction r over n end fraction)nt, where A(t) is the final amount, A0 is the initial investment, r is the growth rate expressed as a decimal, n is the number of compounding periods per year, and t is time in years.

A. $8,519.54
B. $8,505.65
C. $13,458.78
D. $8,580.23

Answer 1

@soprano.h.d0816 @RhondaSommer @DanJS

Answer 2

yeah this is the same sort of exponential function, y=A*b^x

except when you figure the rate of change b on this one, it will only be 1/4 of that total 5.75 yearly percent., that is the r/n thing,

Answer 3

n is the number of compounding cycles each period of t, 1 year

Answer 4

and then instead of just raised to the t power, compounding once per year, it is now n*t, 4*t, since it compounds quarterly, 4 times each year

Answer 5

A. $8,519.54
B. $8,505.65
C. $13,458.78
D. $8,580.23

Answer 6

so its B? @DanJS

Answer 7

this one they give you the desired future value when t=6 years, and ask for the initial investment

\[\large A(t) = A _{0}*(1+\frac{ .0575 }{ 4 })^{4*t}\]

that is the function,

when t=6, you want A(6) = 12000

\[\huge A(6) = 12000= A _{0}*(1+\frac{ .0575 }{ 4 })^{4*t}\]

Answer 8

solve for that initial amount needed Ao,
t=6

Answer 9

okay i got a weird number

Answer 10

\[\huge 12000= A _{0}*(1.014375)^{24}\]

Answer 11

12000/(1.014375)^24

Answer 12

oh so its 8519?

Answer 13

yep

Answer 14

okay and one more question if thats okay

Answer 15

sure

Answer 16

The half-life of a substance is how long it takes for half of the substance to decay or become harmless (for certain radioactive materials). The half-life of a substance is 8.6 days and there is an amount equal to 15 grams now. What is the expression for the amount A(t) that remains after t days, and what is the amount of the substance remaining (rounded to the nearest tenth) after 37 days?

Hint: The exponential equation for half-life is

Answer 17

A. A(t) = 15(0.5)8.6t, 0.0 gram remaining
B. A(t) = 15(0.5)t/8.6, 0.8 gram remaining
C. A(t) = 8.6(15)(0.5)t, 0.0 gram remaining
D. A(t) = 15(0.5)8.6/t, 12.8 grams remaining

Answer 18

i just need this one so i can go to sleep

Answer 19

this is the same form of general exponential
y(t) = A*b^t

Answer 20

so how do i plug it in

Answer 21

The given info is

start amount at t=0 is 15g
half is left after t=8.6

Answer 22

so 8.6 is C?

Answer 23

each of those is a point on the graph, time t=0 and amojnt y(t) = 15 grams

y(0) = 15 * (b)^0
= 15* 1

so A=15, you have

y(t) = 15*b^t

Answer 24

A. A(t) = 15(0.5)8.6t, 0.0 gram remaining
B. A(t) = 15(0.5)t/8.6, 0.8 gram remaining
C. A(t) = 8.6(15)(0.5)t, 0.0 gram remaining
D. A(t) = 15(0.5)8.6/t, 12.8 grams remaining

Answer 25

so its gonna be A right

Answer 26

the other thing is half is left after t=8.6, so when half is gone, y(t)/A will be 1/2 or 0.5

Answer 27

1/2 = b^8.6

that gives you the b value,

Answer 28

so the answer is A?

Answer 29

@DanJS

Answer 30

or C?

Answer 31

I'm sorry i just really wanna finish since I'm timed @DanJS

Answer 32

y(t) = 15*(0.922564)^t

for t=37 days

y(37) = about 0.7 g left

Answer 33

A. A(t) = 15(0.5)8.6t, 0.0 gram remaining
B. A(t) = 15(0.5)t/8.6, 0.8 gram remaining
C. A(t) = 8.6(15)(0.5)t, 0.0 gram remaining
D. A(t) = 15(0.5)8.6/t, 12.8 grams remaining
which answer would it be then

Answer 34

i just need to know the letter for the answer @DanJS

Answer 35

prolly B

Answer 36

they just wrote the thing in a different form

Answer 37

you sure?

Answer 38

yes

Answer 39

thank you it was right

Answer 40

yes, just study up that general exponential y=A*b^x, all the probs were really the same, just different context