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Algebra II : Graphing Data

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Example Questions

Example Question #1 : Interpolations

Find the value of when x=-2 given the points (8,1) and (4, 9) .

Possible Answers:

$\frac{7}{2}$

Correct answer:

Explanation:

Write the interpolation formula.

$y=y_1 + \frac{(x-x_1)(y_2-y_1)}{(x_2-x_1)}$

Identify and substitute the values.

$y=1 + \frac{(-2-8)(9-1)}{(4-8)}$

Simplify the fraction.

$y=1 + \frac{-10(8)}{-4} = 1+\frac{-80}{-4} =1+20 = 21$

The answer is:

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Example Question #31 : Graphing Data

What is the next number in this sequence: 8, 27, 64, 125 ?

Possible Answers:

350

225

125

216

Correct answer:

216

Explanation:

Find the pattern of the sequence:

This pattern is $2^{3}, 3^{3}, 4^{3}, 5^{3}$ so the next number in the sequence would be $6^{3}=216$

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Example Question #2 : Extrapolations

Find the next 2 numbers in this sequence: 33, 46, 72, 111

Possible Answers:

$163\ \textup{and }312$

$163\ \textup{and }228$

$150\ \textup{and }211$

$179\ \textup{and }228$

Correct answer:

$163\ \textup{and }228$

Explanation:

Find the pattern in this sequence of numbers:

In this case, the pattern is adding 13n to the previous number where n= how many numbers came before the current number.

$33+13, 46+(13 \times 2), 72+(13 \times 3)$

so the first number we are looking for would be:

$111+ (13 \times 4)=163$

the second number we are looking for would be:

$163 + (13\times 5)=228$

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Example Question #3 : Extrapolations

The amount of water inside of a leaky boat is measured periodically after the boat has been in the water in different periods of time and are found to have a linear relationship. The results are given in the following chart:

Time in water (mins)	Amount of water in the boat (gal)
0	0
6	4.8
19	15.2
28	22.4

Using the method of linear extrapolation based on the data from the table, how much water would you expect to be in the boat after minutes?

Possible Answers:

$39\:Gal$

$48\:Gal$

$37.725\:Gal$

$42.4\: Gal$

$41.4\:Gal$

Correct answer:

$42.4\: Gal$

Explanation:

To extrapolate the results of the study out to 53 minutes, first we have to determine an equation representing the relationship between time passed and amount of water; we can write our equation in slope-intercept form:

y=mx+b

Where our y-axis represents amount of water and the x-axis represents time. We can pick 2 points and label them Point 1 and Point 2; looking back at the table:

Time in water (mins)	Amount of water in the boat (gal)
0	0
6	4.8
19	15.2
28	22.4

We label Point 1 as (6,4.8) and Point 2 as (19,15.2) ; we plug these points into the slope formula as follows:

$\frac{y_{2}-y_{1}}{x_2-x_1}=\frac{15.2-4.8}{19-6}=\frac{10.4}{13}=0.8$

So, the slope of our line that describes how much water is in the boat is m=0.8 ; to find our term, the y-intercept, we need to pick a point on the graph and plug in our slope to solve for y-intercept. Let's once again choose the point (6,4.8) :

4.8=(0.8)(6)+b

Simplify the expression and we find that b=0, so our slope-intercept equation is:

y=(0.8)x

Plugging in a value of 53 for , we find that:

y=(0.8)(53)=42.4

So the answer is 42.4 gallons of water in the boat after 53 minutes.

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Algebra II : Graphing Data

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Interpolations

Example Question #31 : Graphing Data

Example Question #2 : Extrapolations

Example Question #3 : Extrapolations

All Algebra II Resources

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