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

Total Earnings ( | 60 | 70 | 80 |

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 = 10

Step: 4

Change in input values = 1

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 0 1 = 10

Step: 6

Working backwards with the values in the table, we get (0, 50).

Step: 7

So, the linear equation that satisfies the table is y = 10x + 50.

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

Correct Answer is : y = 10x + 50

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

Sunny bought a truck for $20,000 whose value decreases by 3% each year.

Step: 2

Its value after 2 years should be $18,818.

Step: 3

This is the most appropriate situation in which one quantity changes at a constant rate per unit interval in relation to another.

Correct Answer is : Sunny bought a truck for $20,000. The value of the truck decreases by 3% each year. Its value after 2 years should be $18,818.

Step: 1

The price of gasoline in the year 1995 was $15 per gallon and its price increases by 3% each year.

Step: 2

The value of the gasoline after 3 years should be $16.35.

Step: 3

This is the most appropriate situation in which one quantity changes at a constant rate per unit interval in relation to another.

Correct Answer is : The price of gasoline in the year 1995 was $15 per gallon and its price increases by 3% each year. The value of the gasoline after 3 years should be $16.35.

Step: 1

Revenues of a factory in the subsequent years were 5 billion dollars, 10.3 billion dollars, 21.218 billion dollars, 43.70908 billion dollars and 90.040705 billion dollars.

Step: 2

This is the most appropriate situation in which one quantity changes at a constant rate per unit interval in relation to another.

Correct Answer is : Revenues of a factory in the subsequent years were 5 billion dollars, 10.3 billion dollars, 21.218 billion dollars, 43.70908 billion dollars and 90.040705 billion dollars.

Step: 1

The initial amount is $16000 and the growth rate in every year is $2000, since the year 1990.

Step: 2

So, the situation given in Choice C is the most appropriate situation for the given graph.

Correct Answer is : Sam′s annual income was $16000 in 1990.Since then, every year his income hasincreased by $2000.

Step: 1

Joe deposited $1000 in his account in 2005 and the money grew from $1000 to $8000 in 3 years, at a constant rate.

Step: 2

Correct Answer is : Joe deposited $1000 in his account in 2005 and the money grew from $1000 to $8000 in 3 years, at a constant rate.

Step: 1

Caroline invested $750 in a mutual fund in the year 1998.

Step: 2

Its value increased by $25 in the subsequent years.

Step: 3

Correct Answer is : Caroline invested $750 in a mutual fund in the year 1998. Its value increased by $25 in the subsequent years.

Step: 1

From 1983 through 1989, the number of students in an institute increased by 200.

Step: 2

Correct Answer is : From 1983 through 1989, the number of students in an institute increased by 200.

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