Question 10 of 10 To model some data, Pat ran an exponential regression and found the
regression equation to be y = 0.496.1.3%,
The scatterplot and residual plot are shown.
Scatterplot
Residual plot
y
4
2
Select the true statements.
O A. The model is a good fit because the patterns in the plots are both
curves.
B. The points in the scatterplot follow a line, so a linear model would
fit the data better.
O C. The residual plot changes direction, so a quadratic model would
fit the data better.
D. The model is not a good fit because there is a pattern in the
residual plot.

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.

one way is to plot 2 points and connect them which is what the answer choices seem to be doing

it seems to be using the y intercept and the slope

y=mx+b, m=slope, b=y-intercept (where the line hits the y axis)

slope=rise/run

y-intercept is where the line croses the y axis which is the y value where x=0

you seem to have given the wrong slope (you didn't put a number in front of te x, it should be 2/3 or 3/2 looking at the answer options)

y=? x-1

m=? because of your fault

b=-1

so it has a slope of ? and y intercept of -1, meaning it pases through (0,-1)

so draw point (0,-1) (so answer is one of the first 2 option)

if slope is 2/3, then go up 2 and right 3 then draw line through those 2 points

if slope is 3/2, then go up 3 and right 2 then draw line through those 2 points