if $3000 is invested at 6% simple interest, the initial value and the future value s at the end of each of 5…

if $3000 is invested at 6% simple interest, the initial value and the future value s at the end of each of 5 years is shown in the table below. then answer part (a) through (d).\n|year (t)|0|1|2|3|4|5|\n|future value (s)|3000|3180|3360|3540|3720|3900|\n(a) can a linear function model exactly the points from the table? explain.\na. yes, the first differences are constant for uniform inputs.\nb. no, the first differences are not constant for uniform inputs.\n(b) if so, find a linear function s = f(t) that models the points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. s = 180t + 3000 (type an equation.)\nb. a linear model cannot exactly model the points from the table.\n(c) use the model to find the future value of this investment at the end of the 7th year.\nthe future value of this investment at the end of the 7th year is $□
Answer
Explanation:
Step1: Identify the linear - function formula
The linear - function formula is $S = f(t)=mt + b$, where $m$ is the slope and $b$ is the y - intercept.
Step2: Calculate the slope $m$
The slope $m$ can be calculated using the formula $m=\frac{S_2 - S_1}{t_2 - t_1}$. Taking two points from the table, say $(t_1 = 0,S_1 = 3000)$ and $(t_2 = 1,S_2 = 3180)$. Then $m=\frac{3180 - 3000}{1-0}=180$.
Step3: Determine the y - intercept $b$
When $t = 0$, $S=b$. From the table, when $t = 0$, $S = 3000$, so $b = 3000$. The linear function is $S=180t + 3000$.
Step4: Find the future value at $t = 7$
Substitute $t = 7$ into the function $S=180t + 3000$. Then $S=180\times7+3000=1260 + 3000=4260$.
Answer:
(a) A. Yes, the first differences are constant for uniform inputs (b) A. $S = 180t+3000$ (c) $4260$