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Question
the scatter plot shows the time spent texting, x, and the time spent exercising, y, by each of 23 students last week. use the scatter plot to answer the parts below. (note that you can use the graphing tools to help you approximate the line.)
scratch area (not part of answer)
(a) write an approximate equation of the line of best fit. round the coefficients to the nearest hundredth.
y =
(b) using your equation from part (a), predict the time spent texting for a student who spends 6 hours exercising. round your prediction to the nearest hundredth.
hours
Step1: Identify two points on the line of best fit
Looking at the scatter plot, we can estimate two points. Let's assume two points: (1, 10) and (9, 1) (these are approximate points from the scatter plot's trend).
Step2: Calculate the slope (m)
The formula for slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$. Substituting the points $(x_1, y_1) = (1, 10)$ and $(x_2, y_2) = (9, 1)$:
$m = \frac{1 - 10}{9 - 1} = \frac{-9}{8} = -1.125 \approx -1.13$ (rounded to nearest hundredth)
Step3: Find the y-intercept (b)
Using the point-slope form $y - y_1 = m(x - x_1)$ with point (1, 10) and $m = -1.13$:
$y - 10 = -1.13(x - 1)$
$y - 10 = -1.13x + 1.13$
$y = -1.13x + 11.13$ (this is the approximate line of best fit for part (a))
Step4: Predict for x = 6 (part b)
Substitute $x = 6$ into the equation $y = -1.13x + 11.13$:
$y = -1.13(6) + 11.13$
$y = -6.78 + 11.13$
$y = 4.35$
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(a) The approximate equation of the line of best fit is $y = -1.13x + 11.13$ (answers may vary slightly based on point estimation).
(b) The predicted time spent texting for a student who spends 6 hours exercising is $\boxed{4.35}$ hours.