David invested $7,300 in an account paying an interest rate of 1.6% compounded continuously. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $9,070?

13.6

Step-by-step explanation:

A= Pe^ rt

A=9070   P=7300   t= .016

9070 = 7300e^0.016t

divide by 7300

1.24246575342= e^0.016t

ln(1.24246575342)= ln e^0.016t

ln(1.24246575342)= 0.016t

divide by 0.016

t = 13.6

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.