An investment will pay $100 at the end of each of the next 3 years, $200 at the end of year 4, $300 at the end of year 5, and $500 at the end of year 6. if other investments of equal risk earn 8% annually, what is its present value? its future value?

$1,466.23 Will be your answer mate

Step-by-step explanation:

The present value is calculated by discounting each of the cash flows to the present therefore if we assume no salvage value:  

PV = $100 / 1.08 + $100 / 1.08^2 + $100 / 1.08^3 + $200 / 1.08^4 + $300 / 1.08^5 + $600 / 1.08^6  

PV = $923.98  

Technically, as we have assumed there are no salvage value then in the future value will be zero. If you are asking how much remains in the investment. If you are asking how much the investment has yielded, keep in mind that the cash flows are not by definition reinvested so the time value of those cash flows can't be accounted for. If you do not account for the time value of those cash flows then the future value is just:  

$100 * 3 + $200 + $300 + $500 = $1,300  

However if you use the 8% per annum effective rate as a discount rate to take into consideration the time value of the money which is basically saying that you can reinvest the cash flows at 8% per annum effective then the future value would be:  

$100 * 1.08^5 + $100 * 1.08^4 + $100 * 1.08^3 + $200 * 1.08^2 + $300 * 1.08 + 500 = $1,466.23

you'll get your answer as : $1,466.23 !

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24