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).\nyear (t) 0 1 2 3 4 5\nfuture value (s) 3000 3180 3360 3540 3720 3900\n(a) are the first - differences constant for uniform inputs?\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 $4260.\nis this an interpolation or an extrapolation from the data?\na. extrapolation\nb. interpolation

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).\nyear (t) 0 1 2 3 4 5\nfuture value (s) 3000 3180 3360 3540 3720 3900\n(a) are the first - differences constant for uniform inputs?\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 $4260.\nis this an interpolation or an extrapolation from the data?\na. extrapolation\nb. interpolation

Answer

Explanation:

Step1: Identify the linear - function form

The general form of a linear function is $S = mt + b$, where $m$ is the slope and $b$ is the initial value. The initial investment is $b = 3000$. The slope $m$ can be found by calculating the change in $S$ over the change in $t$. For example, from $t = 0$ to $t = 1$, $S$ changes from $3000$ to $3180$. The change in $S$ ($\Delta S$) is $3180 - 3000=180$, and the change in $t$ ($\Delta t$) is $1 - 0 = 1$. So, $m = 180$. Thus, the linear function is $S=180t + 3000$.

Step2: Calculate the future - value at $t = 7$

Substitute $t = 7$ into the linear function $S = 180t+3000$. $S=180\times7 + 3000$. First, calculate $180\times7=1260$. Then, $S=1260 + 3000=4260$.

Step3: Determine interpolation or extrapolation

The data in the table is for $t$ values from $0$ to $5$. Since $t = 7$ is outside the range of $t$ values in the table, this is an extrapolation.

Answer:

A. Extrapolation