 Solve for a. 1.10 = 2a 2.–5a = –16 ANSWER 5 16 5 3. Write an equation of the line that passes through the points (0, 0) and (4, 8). y = 2x Solve for a.

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Solve for a. 1.10 = 2a 2.–5a = –16 ANSWER 5 16 5 3. Write an equation of the line that passes through the points (0, 0) and (4, 8). y = 2x Solve for a.

Main Concept

Write and graph a direct variation equation EXAMPLE 1 Write and graph a direct variation equation that has (–4, 8) as a solution. Use the given values of x and y to find the constant of variation. Write direct variation equation. y = ax 8 = a(–4) Substitute 8 for y and – 4 for x. –2 = a Solve for a. Substituting –2 for a in y = ax gives the direct variation equation y = –2x.

GUIDED PRACTICE for Example 1 Write and graph a direct variation equation that has the given ordered pair as a solution. 1. (3, –9) ANSWER y = – 3x. y = ax -9 = a(3) -3 = a

GUIDED PRACTICE for Example 1 Write and graph a direct variation equation that has the given ordered pair as a solution. 2. (–7, 4) y = ax 4 = a(-7) -4/7 = a y = -4/7x

Write and apply a model for direct variation EXAMPLE 2 Meteorology Hailstones form when strong updrafts support ice particles high in clouds, where water droplets freeze onto the particles. The diagram shows a hailstone at two different times during its formation.

Write and apply a model for direct variation EXAMPLE 2 a. Write an equation that gives the hailstone’s diameter d (in inches) after t minutes if you assume the diameter varies directly with the time the hailstone takes to form. b. Using your equation from part (a), predict the diameter of the hailstone after 20 minutes. d = atd = at 0.75 = a(12) 0.0625 = a An equation that relates t and d is d = 0.0625t. After t = 20 minutes, the predicted diameter of the hailstone is d = 0.0625(20) = 1.25 inches. d = atd = at

Correlation: +1 Correlation: -1 Correlation: 0 Correlation: +1/2 Correlation: -1/2

Describe correlation EXAMPLE 1 Describe the correlation shown by each scatter plot. Positive CorrelationNegative Correlation The first scatter plot shows a positive correlation, because as the number of cellular phone subscribers increased, the number of cellular service regions tended to increase. The second scatter plot shows a negative correlation, because as the number of cellular phone subscribers increased, corded phone sales tended to decrease. Positive or Negative

EXAMPLE 2 Estimate correlation coefficients Tell whether the correlation coefficient for the data is closest to – 1, – 0.5, 0, 0.5, or 1. a.a. SOLUTION a. The scatter plot shows a clear but fairly weak negative correlation. So, r is between 0 and – 1, but not too close to either one. The best estimate given is r = – 0.5. ( The actual value is r –0.46. )

Estimate correlation coefficients EXAMPLE 2 b. The scatter plot shows approximately no correlation. So, the best estimate given is r = 0. (The actual value is r – 0.02.) SOLUTION b.

Estimate correlation coefficients EXAMPLE 2 c. c. The scatter plot shows a strong positive correlation. So, the best estimate given is r = 1. (The actual value is r 0.98.) SOLUTION

GUIDED PRACTICE for Examples 1 and 2 For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, – 0.5, 0, 0.5, or 1. 1. ANSWER (a) positive correlation (b) r = 0.5

GUIDED PRACTICE for Examples 1 and 2 For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, – 0.5, 0, 0.5, or 1. 2. ANSWER (a) negative correlation (b) r = –1

GUIDED PRACTICE for Examples 1 and 2 For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, –0.5, 0, 0.5, or 1. 3. ANSWER (a) no correlation (b) r = 0

EXAMPLE 3 Approximate a best-fitting line The table shows the number y (in thousands) of alternative-fueled vehicles in use in the United States x years after 1997. Approximate the best-fitting line for the data. x01234567 y280295322395425471511548 Alternative-fueled Vehicles This will be very important for our project next week!

Approximate a best-fitting line EXAMPLE 3 SOLUTION STEP 1 Draw a scatter plot of the data. STEP 2 Sketch the line that appears to best fit the data. One possibility is shown.

Approximate a best-fitting line EXAMPLE 3 STEP 3 Choose two points that appear to lie on the line. For the line shown, you might choose (1,300), which is not an original data point, and (7,548), which is an original data point. STEP 4 Write an equation of the line. First find the slope using the points (1,300) and (7,548). 248 6 m = = 41.3 548 – 300 7 – 1

Approximate a best-fitting line EXAMPLE 3 Use point-slope form to write the equation. Choose (x 1, y 1 ) = (1,300). y – y 1 = m(x – x 1 ) Point-slope form y – 300 = 41.3(x – 1) Substitute for m, x 1, and y 1. Simplify. y 41.3x + 259 ANSWER An approximation of the best-fitting line is y = 41.3x + 259.

EXAMPLE 4 Use a line of fit to make a prediction Use the equation of the line of fit from Example 3 to predict the number of alternative-fueled vehicles in use in the United States in 2010. SOLUTION Because 2010 is 13 years after 1997, substitute 13 for x in the equation from Example 3. y = 41.3x + 259 = 41.3(13) + 259 796

Use a line of fit to make a prediction EXAMPLE 4 ANSWER You can predict that there will be about 796,000 alternative-fueled vehicles in use in the United States in 2010.

Classwork: Classwork: Worksheet 2-5 (1-21 odd) Worksheet 2-6 (1-10 all)

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