Year | Number of girls |

1998 | 2400 |

1999 | 3000 |

Step: 1

Average rate of change = C h a n g e i n t h e n u m b e r o f g i r l s a d m i t t e d i n s c h o o l C h a n g e i n t i m e

Step: 2

= 3 0 0 0 - 2 4 0 0 1 9 9 9 - 1 9 9 8

Step: 3

= 600 per year

Step: 4

So, the number of girls admitted in the year 2003 is 2400 + 5(600) = 5400

Correct Answer is : 5400

Step: 1

Average rate of change = C h a n g e i n t h e a n n u a l i n c o m e C h a n g e i n t i m e

Step: 2

= 3 6 0 0 0 - 3 0 0 0 0 1 9 9 8 - 1 9 9 6

Step: 3

= 6 0 0 0 2 = $3000 per year

Step: 4

So, the annual income of Jeff in the year 2004 is $(30,000 + 8(3,000)) = $54,000

Correct Answer is : $54,000

Number of Minutes ( | 10 | 20 | 30 | 40 | 50 |

Cost in Dollars ( | 11.50 | 13 | 14.50 | 16 | 17.50 |

Step: 1

Linear equation in intercept form is y = a + b x , where b is the rate of change and a is the y - intercept.

Step: 2

The input variable x is the number of minutes and the output variable y gives the total cost.

Step: 3

Change in output values = 1.50.

Step: 4

Change in input values =10.

Step: 5

Rate of change= C h a n g e i n o u t p u t v a l u e s C h a n g e i n i n p u t v a l u e s = 1 . 5 0 1 0 = 0.15

Step: 6

Working backwards with the values in the table, we get (0, 10).

Step: 7

So, the linear equation that satisfies the table is y = 0.15x +10.

[Substitute rate of change and y - intercept values.]

Correct Answer is : y = 0.15x + 10

Number of Minutes ( | 30 | 60 | 90 | 120 | 150 |

Water Level ( | 40 | 50 | 60 | 70 | 80 |

Number of Extra Hours ( | 1 | 2 | 3 |

Total Earnings ( | 90 | 110 | 130 |

Step: 1

Linear equation in intercept form is y = a + b x , where b is the rate of change and a is the y - intercept.

Step: 2

The input variable x is the number of extra hours and the output variable y gives the total earnings of John per day.

Step: 3

Change in output values = 20

Step: 4

Change in input values = 1

Step: 5

Rate of change = c h n a g e i n o u t p u t v a l u e s c h a n g e i n i n p u t v a l u e s = 2 0 1 = 20

Step: 6

Working backwards with the values in the table, we get (0, 70).

Step: 7

So, the linear equation that satisfies the table is y = 20x + 70.

[Substitute rate of change and y - intercept values.]

Correct Answer is : y = 20x + 70

Sales Worth ( | 100 | 200 | 300 | 400 |

Expenditure in Dollars ( | 107 | 114 | 121 | 128 |

Step: 1

Linear equation in intercept form is y = a + b x , where b is the rate of change and a is the y - intercept.

Step: 2

The input variable x is the sales worth and the output variable y is the expenditure.

Step: 3

Change in output values = 7

Step: 4

Change in input values = 100

Step: 5

Rate of change = c h a n g e i n o u t p u t v a l u e s c h a n g e i n i n p u t v a l u e s = 7 1 0 0 = 0.07

Step: 6

Working backwards with the values in the table, we get (0, 100).

Step: 7

So, the linear equation that satisfies the table is y = 0.07x + 100.

[Substitute rate of change and y - intercept values.]

Correct Answer is : y = 0.07x + 100

Step: 1

The slope of the line = r i s e r u n = $ h o u r =y_{2}-y_{1} x_{2}-x_{1} = 3 2 - 2 4 4 - 3 , where any of the points on the line could be selected.

Step: 2

Step: 3

Since the slope of the line is 8, Sam earns $8 per hour.

Correct Answer is : Sam earns $8 per hour.

Step: 1

The slope of the line = r i s e r u n = $ h o u r = y_{2}-y_{1} x_{2}-x_{1} = 4 . 2 5 - 3 . 5 0 4 - 3 , where any of the points on the line could be selected.

Step: 2

Step: 3

Since the slope of the line is 0.75, the parking rate increases $0.75 per hour.

Correct Answer is : Parking rate for every additional hour increases by $0.75.

Step: 1

The rate of increase is 150 per year.

Step: 2

Let t = 0 represent the year 1985.

Step: 3

The year 1991 is represented by t = 6.

Step: 4

This situation is represented by the linear equation y = 150t + 1000.

Step: 5

At t = 6, y = 150(6) + 1000 = 900 + 1000 = 1900

Step: 6

So, the number of schools in the year 1991 is 1900.

Correct Answer is : 1900

Input | - 1.2 | 0 | 0.6 | 1.5 | 2.4 | 3.3 | 3.6 |

Output | 5.2 | 4 | 3.4 | 2.5 | 1.6 | 0.7 | 0.4 |

Step: 1

Rate of change = c h a n g e i n o u t p u t v a l u e s c h a n g e i n i n p u t v a l u e s

Step: 2

Input | Output | Change in input values | Change in output values | Rate of change |

- 1.2 | 5.2 | |||

0 | 4 | 1.2 | -1.2 | - 1 |

0.6 | 3.4 | 0.6 | - 0.6 | - 1 |

1.5 | 2.5 | 0.9 | - 0.9 | - 1 |

2.4 | 1.6 | 0.9 | - 0.9 | - 1 |

3.3 | 0.7 | 0.9 | -0.9 | - 1 |

3.6 | 0.4 | 0.3 | - 0.3 | - 1 |

Step: 3

The rate of change = - 1

Step: 4

The linear equation that represents the data is y = 4 + (- 1) x .

Step: 5

Step: 6

When x = 4.46, y = - 0.46

Correct Answer is : - 0.46

Cell phones | 1 | 2 | 3 |

Earnings | 72 | 80 | 88 |

Step: 1

Rate of change = 8 0 - 7 2 2 - 1 = 8

Step: 2

This situation can be modeled by the linear equation y = 8t + 64

Step: 3

96 = 8t + 64 gives t = 4

Step: 4

Joseph needs to sell 4 cell phones.

Correct Answer is : 4

Months | 24 | 27 | 30 | 33 |

Weight (pounds) | 30 | 31 | 32 | 33 |

Step: 1

Rate of change = C h a n g e i n w e i g h t C h a n g e i n t i m e

Step: 2

= 31-30 27-24 = 1 3 pounds per month

Step: 3

So, the rate of change in weight of a child is 1 3 × 12 = 4 pounds per year.

Correct Answer is : 4 pounds per year

- Determining Whether a given Relation is a Function-Gr 8-Solved Examples
- Classifying Relations as Linear or Nonlinear-Gr 8-Solved Examples
- Identifying Functions-Gr 8-Solved Examples
- Graphing Linear Functions for a Given Domain-Gr 8-Solved Examples
- Graphing Linear Functions based on an x/y Table-Gr 8-Solved Examples
- Graphing Linear Equations-Gr 8-Solved Examples
- Finding the Rule of a Function Table-Gr 8-Solved Examples
- Representing Exponential Functions Using Tables or Graphs-Gr 8-Solved Examples

- Rate Of Change
- Slope