ctive 4-3 : use line of best fit to make predictions.
what trend line is appropriate to represent the scatter plot?
a. y = .75x + 30
b. y = 1.5x - 28
c. y = 15x + 30
d. y = -10x + 35
2. create a scatter plot for the data shown in the table representing the number of minutes spent walking on a treadmill and the number of calories burned.
| time | 18 | 24 | 30 | 40 | 42 | 48 | 52 |
| calories | 260 | 280 | 320 | 380 | 400 | 440 | 475 |
3. sketch a line of best fit for the data. predict what the amount of calories that would be burned when walking for 35 minutes.
4. use your calculator to find the line of best fit for the given data showing the temperature in degrees celsius and the corresponding temperature in degrees fahrenheit. using that equation, what temperature in degrees fahrenheit, would be 17 degrees celsius be?
| celcius | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
| fahrenheit | 32 | 41 | 50 | 59 | 68 | 77 | 86 |
part a: line of best fit equation:
part b: prediction for 17 degrees celsius:

Question

ctive 4-3 : use line of best fit to make predictions.
what trend line is appropriate to represent the scatter plot?
 a. y = .75x + 30
 b. y = 1.5x - 28
 c. y = 15x + 30
 d. y = -10x + 35
2. create a scatter plot for the data shown in the table representing the number of minutes spent walking on a treadmill and the number of calories burned.
| time | 18 | 24 | 30 | 40 | 42 | 48 | 52 |
| calories | 260 | 280 | 320 | 380 | 400 | 440 | 475 |
3. sketch a line of best fit for the data. predict what the amount of calories that would be burned when walking for 35 minutes.
4. use your calculator to find the line of best fit for the given data showing the temperature in degrees celsius and the corresponding temperature in degrees fahrenheit. using that equation, what temperature in degrees fahrenheit, would be 17 degrees celsius be?
| celcius | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
| fahrenheit | 32 | 41 | 50 | 59 | 68 | 77 | 86 |
part a: line of best fit equation:
part b: prediction for 17 degrees celsius:

Accepted Answer

### Problem 1 (Trend Line for Scatter Plot)
# Explanation:
## Step 1: Analyze the scatter plot's slope and y-intercept
The scatter plot has a positive slope (upward trend) and a y-intercept around 30. Let's check the options:
- Option a: $ Y = 0.75x + 30 $ – positive slope, y-intercept 30.
- Option b: $ Y = 1.5x - 28 $ – y-intercept -28 (too low, doesn't match plot's y-intercept ~30).
- Option c: $ Y = 15x + 30 $ – slope 15 is too steep (plot's slope is gentle).
- Option d: $ Y = -10x + 35 $ – negative slope (plot has positive slope, so eliminate).

## Step 2: Confirm the best fit
The plot's slope is gentle (less than 1.5) and y-intercept ~30. So $ Y = 0.75x + 30 $ (option a) is appropriate. Wait, but the original marked option is c, but based on the plot, let's recheck. Wait, maybe the grid: if x is small (like 0, y=30; x=10, y≈37.5 (0.75*10+30=37.5), which matches the plot. Option c: 15x+30 at x=10 would be 180, which is way too high. So likely a typo, but based on analysis, a is correct. Wait, maybe the plot's x-axis: if x is 0, y=30; x=20, y=0.75*20+30=45, which matches the plot (dots around x=20, y=45). So option a is correct.

# Answer:
a. $ Y = 0.75x + 30 $

### Problem 2 (Create Scatter Plot)
# Explanation:
## Step 1: Identify variables
Time (x-axis: 18,24,30,40,42,48,52) and Calories (y-axis: 260,280,320,380,400,440,475).

## Step 2: Plot points
For each (Time, Calories) pair: (18,260), (24,280), (30,320), (40,380), (42,400), (48,440), (52,475). Mark these on a grid with x as Time (0 - 60) and y as Calories (0 - 500).

# Answer:
Scatter plot with points at (18,260), (24,280), (30,320), (40,380), (42,400), (48,440), (52,475) on a grid with x=Time, y=Calories.

### Problem 3 (Predict Calories at 35 minutes)
# Explanation:
## Step 1: Use the line of best fit (given $ y = 18x + 260 $ – wait, let's recalculate the correct line of best fit. Let's find slope (m) and y-intercept (b) for Time (x) and Calories (y):

Data: (18,260), (24,280), (30,320), (40,380), (42,400), (48,440), (52,475)

Slope $ m = \frac{\Delta y}{\Delta x} $. Take (18,260) and (52,475): $ \Delta y = 475 - 260 = 215 $, $ \Delta x = 52 - 18 = 34 $, $ m ≈ 6.32 $. Wait, the given $ y = 18x + 260 $ is wrong. Wait, maybe the student's mistake. Let's calculate correctly.

Using linear regression:

$ \bar{x} = \frac{18+24+30+40+42+48+52}{7} = \frac{254}{7} ≈ 36.29 $

$ \bar{y} = \frac{260+280+320+380+400+440+475}{7} = \frac{2555}{7} ≈ 365 $

$ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} $

Calculate (x_i - \bar{x})(y_i - \bar{y}):

(18-36.29)(260-365) ≈ (-18.29)(-105) ≈ 1920.45

(24-36.29)(280-365) ≈ (-12.29)(-85) ≈ 1044.65

(30-36.29)(320-365) ≈ (-6.29)(-45) ≈ 283.05

(40-36.29)(380-365) ≈ (3.71)(15) ≈ 55.65

(42-36.29)(400-365) ≈ (5.71)(35) ≈ 199.85

(48-36.29)(440-365) ≈ (11.71)(75) ≈ 878.25

(52-36.29)(475-365) ≈ (15.71)(115) ≈ 1806.65

Sum ≈ 1920.45+1044.65=3065.1+283.05=3348.15+55.65=3403.8+199.85=3603.65+878.25=4481.9+1806.65=6288.55

$ \sum (x_i - \bar{x})^2 $:

(18-36.29)^2 ≈ 334.5

(24-36.29)^2 ≈ 151.0

(30-36.29)^2 ≈ 39.5

(40-36.29)^2 ≈ 13.8

(42-36.29)^2 ≈ 32.6

(48-36.29)^2 ≈ 137.1

(52-36.29)^2 ≈ 246.8

Sum ≈ 334.5+151=485.5+39.5=525+13.8=538.8+32.6=571.4+137.1=708.5+246.8=955.3

$ m ≈ \frac{6288.55}{955.3} ≈ 6.58 $

$ b = \bar{y} - m\bar{x} ≈ 365 - 6.58*36.29 ≈ 365 - 238 ≈ 127 $

So line of best fit: $ y ≈ 6.58x + 127 $

At x=35: $ y ≈ 6.58*35 + 127 ≈ 230.3 + 127 ≈ 357.3 $

But the student used $ y = 18x + 260 $, which at x=35 is 18*35 + 260 = 630 + 260 = 890 (wrong, too high). So correct prediction is ~357.

# Answer:
Using correct linear regression, ~357 calories (or ~360). Using the student's incorrect line, 890 (but that's wrong). The correct prediction is approximately 357 calories.

### Problem 4 (Line of Best Fit and Prediction)
#### Part A: Line of Best Fit Equation
# Explanation:
## Step 1: Identify variables
Celsius (x: 0,5,10,15,20,25,30) and Fahrenheit (y: 32,41,50,59,68,77,86)

## Step 2: Calculate slope (m)
$ m = \frac{\Delta y}{\Delta x} = \frac{41-32}{5-0} = \frac{9}{5} = 1.8 $? Wait, no: 50-41=9, 10-5=5, so slope is 9/5=1.8? Wait, 32 to 41: +9 (x+5), 41 to 50: +9 (x+5), so slope is 9/5=1.8? Wait, 32 (x=0), 41 (x=5): 41= m*5 + b. At x=0, y=32, so b=32. Then 41=5m +32 → 5m=9 → m=1.8 (9/5=1.8). So line of best fit: $ y = 1.8x + 32 $ (which is the actual Celsius-Fahrenheit formula: $ F = 1.8C + 32 $)

The student's answer $ y = 5x + 41 $ is wrong (at x=0, y=41≠32). Correct line: $ y = 1.8x + 32 $ (or $ y = \frac{9}{5}x + 32 $)

## Step 3: Confirm with linear regression
$ \bar{x} = \frac{0+5+10+15+20+25+30}{7} = 15 $

$ \bar{y} = \frac{32+41+50+59+68+77+86}{7} = \frac{413}{7} = 59 $

$ m = \frac{\sum (x_i - 15)(y_i - 59)}{\sum (x_i - 15)^2} $

Calculate (x_i -15)(y_i -59):

(0-15)(32-59)=(-15)(-27)=405

(5-15)(41-59)=(-10)(-18)=180

(10-15)(50-59)=(-5)(-9)=45

(15-15)(59-59)=0

(20-15)(68-59)=(5)(9)=45

(25-15)(77-59)=(10)(18)=180

(30-15)(86-59)=(15)(27)=405

Sum: 405+180=585+45=630+0=630+45=675+180=855+405=1260

$ \sum (x_i -15)^2 $:

(0-15)^2=225

(5-15)^2=100

(10-15)^2=25

(15-15)^2=0

(20-15)^2=25

(25-15)^2=100

(30-15)^2=225

Sum: 225+100=325+25=350+0=350+25=375+100=475+225=700

$ m = \frac{1260}{700} = 1.8 $ (9/5)

$ b = \bar{y} - m\bar{x} = 59 - 1.8*15 = 59 - 27 = 32 $

So line of best fit: $ y = 1.8x + 32 $ (or $ y = \frac{9}{5}x + 32 $)

# Answer (Part A):
$ y = 1.8x + 32 $ (or $ y = \frac{9}{5}x + 32 $)

#### Part B: Prediction for 17°C
# Explanation:
## Step 1: Use the correct line of best fit
$ y = 1.8x + 32 $. Substitute x=17:

$ y = 1.8*17 + 32 = 30.6 + 32 = 62.6 $ (≈63°F)

The student's answer $ y = 17x + 62 $ is wrong (at x=17, 17*17+62=289+62=351, way too high).

## Step 2: Confirm with the formula
Actual formula: $ F = 1.8C + 32 $, so at C=17, $ F = 1.8*17 + 32 = 30.6 + 32 = 62.6 ≈ 63°F $

# Answer (Part B):
Approximately 63°F (or 62.6°F)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1 (Trend Line for Scatter Plot)

Step 1: Analyze the scatter plot's slope and y-intercept

Step 2: Confirm the best fit

Step 1: Identify variables

Step 2: Plot points

Step 1: Use the line of best fit (given \( y = 18x + 260 \) – wait, let's recalculate the correct line of best fit. Let's find slope (m) and y-intercept (b) for Time (x) and Calories (y):

Answer:

Problem 2 (Create Scatter Plot)