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WGU Applied-Algebra Exam Syllabus Topics:

Section	Objectives
Topic 1: Foundations of Algebra	- Equations and Inequalities 1. Linear equations in one variable 2. Inequalities and interval notation - Algebraic Expressions 1. Order of operations 2. Simplifying expressions 3. Evaluating expressions
Topic 2: Polynomial and Rational Expressions	- Polynomials 1. Factoring techniques 2. Operations with polynomials - Rational Expressions 1. Restrictions on variables 2. Simplification
Topic 3: Exponential and Logarithmic Concepts	- Exponential Functions 1. Exponential equations 2. Growth and decay models - Logarithmic Basics 1. Logarithm properties 2. Conversion between exponential and logarithmic form
Topic 4: Functions and Graphs	- Function Concepts 1. Function notation 2. Domain and range - Graphing Functions 1. Coordinate plane interpretation 2. Transformations of graphs
Topic 5: Linear Functions and Systems	- Systems of Equations 1. Substitution method 2. Elimination method - Linear Equations 1. Point-slope form 2. Slope-intercept form

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WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q41-Q46):

NEW QUESTION # 41
The data in the scatterplot represents the number of monthly train crossings at a particular intersection over time.

Which type of function should be used to model the data?

A. Polynomial
B. Logistic
C. Linear
D. Exponential

Answer: D

Explanation:
The scatterplot shows a decreasing pattern.
At first, the number of monthly crossings decreases quickly. Then the values begin to level off.
This type of pattern is characteristic of an exponential decay model.
A linear model would show points decreasing at a constant rate, forming an approximately straight line. Here, the decrease is not constant; it is steep at first and then slows down.
A logistic model usually has an S-shaped pattern, which is not shown here.
A polynomial model may curve, but the long-term leveling behavior shown in the scatterplot is best represented by an exponential decay function.
Therefore, the correct answer is:
# ( " D " )

NEW QUESTION # 42
A researcher collected data on the average selling price for tickets to a concert over time. The results are shown in the scatterplot. A regression function is graphed with r
2=0.56. The predicted average selling price for tickets after 10.8 days is $167.20.

Is this prediction appropriate?

A. No. The r2value indicates a moderate fit, but x=10.8 is more than 25% of the range beyond the maximum value.
B. Yes. The r2 value indicates a moderate fit, and x=10.8 is within 25% of the range of the maximum value.
C. Yes. The r2 value indicates a strong fit, and x=10.8 is within 50% of the range of the maximum value.
D. No. The r2 value indicates a strong fit, but x=10.8 is more than 50% of the range beyond the maximum value.

Answer: B

Explanation:
The regression model has r2=0.56, which indicates a moderate fit. A moderate fit is not as reliable as a strong fit, but it can still be used for limited prediction when the extrapolation is close to the observed data. The prediction is made at x=10.8 days. This value is slightly beyond the maximum observed x-value, so the prediction is an extrapolation. For a moderate fit, the x-value should remain within about 25% of the range beyond the maximum value. In this case, x=10.8 is within that acceptable range, so the prediction is appropriate. Therefore, the correct answer is B.

NEW QUESTION # 43
Based on collected data, the value of a painting can be modeled using the exponential function f(x)=240(1.12) x
. In this case, x represents the number of years since 2005, and f(x) represents the value of the painting in dollars.
Which value represents the average yearly rate of change of the painting's value from 2013 to 2015?

A. 75.59
B. 74.42
C. 151.17
D. 669.82

Answer: A

Explanation:
This problem asks for an average yearly rate of change over a time interval. Since x represents years since
2005, the year 2013 corresponds to x=8, and the year 2015 corresponds to x=10. Use the average rate of change formula:
10#8
f(10)#f(8)
Evaluating the model gives f(8)=240(1.12)
8
#594.23 and f(10)=240(1.12)
10
#745.40. The total change is approximately 151.17 dollars over 2 years. Dividing by 2 gives about 75.59 dollars per year. Since the question asks for the average yearly rate, not the total increase, the correct answer is B.

NEW QUESTION # 44
The figure shows the graph of p(t), which represents the number of people, p, at a gym thours after 7:00 a.m.

How should the concavity between t=1.3and t=3.4be interpreted?

A. The number of people is decreasing slower and slower.
B. The number of people is increasing slower and slower.
C. The number of people is decreasing faster and faster.
D. The number of people is increasing faster and faster.

Answer: B

Explanation:
The graph represents:
p(t)= " number of people at the gym "
where:
t= " hours after 7:00 a.m. "
The question asks about the concavity of the graph between:
t=1.3
and
t=3.4
On this interval, the graph is still going upward, so the number of people is increasing.
However, the graph is beginning to flatten as it approaches its highest point. That means the slope is positive, but the slope is getting smaller.
In other words:
" The number of people is increasing, but at a slower rate. "
This is called concave down behavior. For a concave down graph that is increasing, the correct interpretation is:
" increasing slower and slower "
Therefore, the correct answer is:
# ( " C " )

NEW QUESTION # 45
The population of fish in a lake is changing according to the function
P(t)=24t+185
where tis the number of months since the beginning of the year and P(t)is the fish population at time t.
Which interpretation of the rate of change is correct?

A. The number of fish in the lake is increasing at a constant rate of 185 fish per month.
B. The number of fish in the lake is decreasing at a constant rate of 24 fish per month.
C. The number of fish in the lake is increasing at a constant rate of 24 fish per month.
D. The number of fish in the lake is decreasing at a constant rate of 185 fish per month.

Answer: C

NEW QUESTION # 46
......

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