suppose the average starting salary for a bachelors degree recipient in the u.s. was $45,609.00 in 2014, and…

suppose the average starting salary for a bachelors degree recipient in the u.s. was $45,609.00 in 2014, and in 2018 it increased to $51,074.00.\na) let y be the average starting salary and t be the number of years since 2014. using these variables, which linear model represents the average starting salary?\n$y = $1366.25t + $45,609\n$y = ($45,609 - $51,074)t\n$y = 1366.25cdot t + $51,074\n$y=\frac{$45,609}{$51,074}t$\ncorrect. good job!\nb) when will the average starting salary reach $75,000.00? round your answer up to the nearest year.
Answer
Explanation:
Step1: Recall linear - equation form
The general form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. In the context of the problem, $y$ is the average starting salary, $t$ is the number of years since 2014. The initial salary (when $t = 0$, i.e., in 2014) is $b=$45,609$.
Step2: Calculate the slope
The change in salary from 2014 to 2018 ($t = 4$ years) is $\Delta y=51074 - 45609=$5465$. The slope $m=\frac{\Delta y}{\Delta t}=\frac{5465}{4}=$1366.25$ per year. So the linear model is $y = 1366.25t+45609$.
Step3: Solve for $t$ when $y = 75000$
Set $y = 75000$ in the equation $y = 1366.25t+45609$. Then $75000=1366.25t + 45609$. Rearrange to get $1366.25t=75000 - 45609=29391$. So $t=\frac{29391}{1366.25}\approx21.51$. Rounding up, $t = 22$ years. Since $t$ is the number of years since 2014, the year is $2014 + 22=2036$.
Answer:
2036