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Algebra II : Algebra II

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Example Questions

Example Question #433 : Basic Statistics

If varies indirectly with $\sqrt{x}$ and when x=4 , y=12.3 due to the effect of a constant, what is the value of when x=17 ?

Possible Answers:

y=5.97

y=25.36

y=50.71

y=16.49

y=0.24

Correct answer:

y=5.97

Explanation:

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when y=12.3 and x=4 .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

K=24.6

2. Use the equation you found to solve for when x=17 .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

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Example Question #2 : Variable Relationships

varies directly with $x^{2}$ . If x=5, y=75 , what is if x=10 ?

Possible Answers:

250

150

300

100

Correct answer:

300

Explanation:

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

75=K(25)

K=3

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

y=300

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Example Question #4 : Identifying Variable Relationships

varies inversely with $x^{3}$ . If x=2, y=2 , then what is equal to when x=5 ?

Possible Answers:

0.654

0.128

0.226

0.197

Correct answer:

0.128

Explanation:

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

$2=\frac{K}{8}$

K=16

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

y=0.128

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Example Question #5 : Identifying Variable Relationships

varies directly with $\sqrt[3]{x}$ and when x=8, y=3 . What is when x=27 ?

Possible Answers:

4.5

5.5

Correct answer:

4.5

Explanation:

1. Since varies directly with $\sqrt[3]{x}$ :

$y=K\sqrt[3]{x}$

2. Use the values given for x and y to solve for K:

$y=K\sqrt[3]{x}$

$3=K\sqrt[3]{8}$

3=K(2)

$K=\frac{3}{2}$

3. Use your new equation with the K you solved for to solve for y when x=27:

$y=(\frac{3}{2})\sqrt[3]{27}$

$y=(\frac{3}{2})(3)=4.5$

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Example Question #441 : Basic Statistics

varies inversely with $\sqrt{x}$ . When $x=4,\ y=0.5$ . What is the value of when x=100 ?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{4}$

$\frac{1}{20}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

1. Since y varies indirectly with $\sqrt{x}$ :

$y=\frac{K}{\sqrt{x}}$

2. Solve for K using the x and y values given:

$0.5=\frac{K}{\sqrt{4}}$

$0.5=\frac{K}{2}$

K=1

3. Using the equation you created by solving for K, find y when x=100:

$y=\frac{1}{\sqrt{x}}$

$y=\frac{1}{\sqrt{100}}$

$y=\frac{1}{10}$

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Example Question #1 : Interpolations

Given the two following points, use interpolation to determine the best estimate for the value f(24.75)

(24,81) , (25,93)

Possible Answers:

82.5

91.5

Correct answer:

Explanation:

Using our two known points, we can use interpolation to determine the value at any point between them with the following formula:

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

Where (x_1,y_1) is our first given point, (x_2,y_2) is our second given point, and (x,y) is the point we want to find. We know our two given points, as well as the x value of our unknown point, so now all we must do is plug in all of our known values and solve for y, our only unknown:

$\frac{93-81}{25-24}=\frac{y-81}{24.75-24}$

$12=\frac{y-81}{0.75}$

$y-81=9\rightarrow y=f(24.75)=90$

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Example Question #1 : Interpolations

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if 110 workers are present.

Linear_interp

Possible Answers:

$1400\textup{ units}$

$1375\textup{ units}$

$1370\textup{ units}$

$1335\textup{ units}$

$1380\textup{ units}$

Correct answer:

$1400\textup{ units}$

Explanation:

We want to do a linear interpolation since the relationship between workers and units can be assumed to be linear. This means there is a constant slope between the points, so the slope between two known points will be equal to the slope between the point we are trying to find and some known point. This is expressed in the relation:

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and (x_1,y_1) and (x_0,y_0) are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

(100,1300) and (200,2300) .

Plugging these into our interpolation formula and knowing x=110 , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

y=1400 .

So there are 1400 units produced when the number of workers is 110 .

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Example Question #1 : Interpolations

Given the points $\left ( 30,51\right )$ and $\left ( 20,36\right )$ , use linear interpolation to find the value of when x=26.5 .

Possible Answers:

44.60

42.38

45.75

49.99

Correct answer:

45.75

Explanation:

Use the formula for interpolation to determine the value of y:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 51) as our x2 and y2 and (20, 36) as our x1 and y1 and we will solve for y using 26.5 for x.

$\frac{51-36}{30-20}=\frac{y-36}{26.5-20}$

$\frac{15}{10}=\frac{y-36}{6.5}$

$\frac{3}{2}(6.5)=y-36$

y=45.75

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Example Question #1 : Interpolations

Given $\left ( 15,10\right )$ and $\left ( 30,15\right )$ use linear interpolation to find when x=17.9 .

Possible Answers:

10.54

10.97

11.34

11.21

Correct answer:

10.97

Explanation:

Use the formula for interpolation:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 15) as x2 and y2, (15, 10) as x1 and y1, and solve for y when x=17.9:

$\frac{15-10}{30-15}=\frac{y-10}{17.9-15}$

$\frac{5}{15}=\frac{y-10}{2.9}$

$(\frac{5}{15})(2.9)=y-10$

y=10.97

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Example Question #5 : Interpolations

Mary measures her height every year on her birthday, starting at 11 until she turns 16. She wants to make a table with all the information gathered, but discovers she lost the piece of paper on which she wrote her height down on her 14^th birthday. Her incomplete table looks like this:

Age (years)	Height (inches)
11	47.5
12	50.25
13	53
14	?
15	58.5
16	61.25

Using the method of linear interpolation, which of the following is the closest estimate of Mary's height on her 14^th birthday?

Possible Answers:

$55.75\:in$

$57\ in$

$53.\bar{3}\:in$

$56\:in$

Not enough information given in the problem.

Correct answer:

$55.75\:in$

Explanation:

Using linear interpolation means that we draw a line between the points on our data set and use that line to estimate a value that lies between two data points; in this case, we have the data from Mary's 13^thand 15^th birthdays, so we can describe a line between those two points and estimate her height at 14. Our line will be written in slope-intercept form:

y=mx+b

Where the variable represents Mary's Age in years and the variable represents her height in inches. First, we need to find the slope. Using 2 points on our table (13,53) and (15,58.5) as point 1 and point 2, respectively, we plug these values into our slope formula:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{(58.5-53)}{(15-13)}=\frac{5.5}{2}=2.75$

Next, we find the y-intercept by plugging in our slope (which we just found) and a point from our table (we'll stick with (13,53) ) and solving for :

53=(2.75)(13)+b

Simplify:

53=35.75+b

Subtract 35.75 from both sides to solve for :

b=17.25

The equation of our interpolation line is:

y=(2.75)x+17.25

So, to get an estimate of Mary's height on her 14^th birthday, we plug in x=14 and solve for :

y=(2.75)(14)+17.25=38.5+17.25=55.75

Our estimate of Mary's height at x=14 is $55.75\:in$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #433 : Basic Statistics

Example Question #2 : Variable Relationships

Example Question #4 : Identifying Variable Relationships

Example Question #5 : Identifying Variable Relationships

Example Question #441 : Basic Statistics

Example Question #1 : Interpolations

Example Question #1 : Interpolations

Example Question #1 : Interpolations

Example Question #1 : Interpolations

Example Question #5 : Interpolations

All Algebra II Resources

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