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WGU Applied-Algebra Exam Syllabus Topics:

Section	Weight	Objectives
Exponents, Radicals, and Quadratic Relationships	15%	- Simplifying radical expressions - Properties of exponents - Solving quadratic equations - Basic quadratic graphs
Algebraic Expressions and Operations	20%	- Order of operations - Simplifying and evaluating expressions - Variables, constants, and coefficients - Operations with polynomials
Systems of Equations and Inequalities	15%	- Solving by substitution and elimination - Graphical solutions - Applications of systems
Graphing and Functions	25%	- Function notation and evaluation - Linear functions and their graphs - Coordinate plane and plotting points - Slope and equations of lines
Linear Equations and Inequalities	25%	- Solving and graphing inequalities - Real-world applications - Solving single-variable equations

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WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q29-Q34):

NEW QUESTION # 29
The growth of an animal population is shown in the graph. The instantaneous rate of change at point Pis 3.74.

Which interpretation of the instantaneous rate of change is correct?

A. In 2007, the population was increasing by 3,740 animals per year.
B. In 2012, the population was increasing by 3,740 animals per year.
C. In 2012, the population was increasing by 3.74 animals per year.
D. In 2007, the population was increasing by 3.74 animals per year.

Answer: A

Explanation:
The graph shows population in thousands, and the horizontal axis represents:
" years since 1995 "
Point Pis located at approximately:
x=12
Since xrepresents years since 1995:
1995+12=2007
So point Pcorresponds to the year:
2007
The instantaneous rate of change at point Pis:
3.74
Because the vertical axis is measured in thousands, this means:
3.74 " thousand animals per year "
Convert thousands to actual animals:
3.74×1000=3740
So in 2007, the population was increasing by approximately:
3,740 " animals per year "

NEW QUESTION # 30
The graphed function n(t)represents the number of people, n, at a park thours after 8:00 a.m. The plotted points Aand Bhave coordinates (2#110)and (6#182).

Which statement gives the correct interpretation of the average rate of change of the number of people over the interval from point Ato point B?

A. The number of people at the park increases at an average rate of 9 per hour.
B. The number of people at the park decreases at an average rate of 18 per hour.
C. The number of people at the park decreases at an average rate of 9 per hour.
D. The number of people at the park increases at an average rate of 18 per hour.

Answer: D

Explanation:
The average rate of change measures how much the output changes per one unit of input.
Here:
t= " hours after 8:00 a.m. "
and
n(t)= " number of people at the park "
The two given points are:
A=(2,110)
and
B=(6,182)
This means:
At t=2hours after 8:00 a.m., there were 110people at the park.
At t=6hours after 8:00 a.m., there were 182people at the park.
Now calculate the average rate of change:
" Average rate of change " =(182-110)/(6-2)
=72/4
=18
Because the value is positive, the number of people is increasing.
So the correct interpretation is:
" The number of people at the park increases at an average rate of 18 people per hour. "

NEW QUESTION # 31
Which statement correctly interprets u(8), given that u(t)=0.21t
3
#4.09t
2
+18.89t+69.5 represents the number of units, u, in an inventory warehouse t hours after 9:00 a.m.?

A. There are approximately 66 units in the inventory at 10:00 a.m.
B. There are approximately 160 units in the inventory at 10:00 a.m.
C. There are approximately 160 units in the inventory at 5:00 p.m.
D. There are approximately 66 units in the inventory at 5:00 p.m.

Answer: D

Explanation:
The function u(t) gives the number of units in the inventory warehouse t hours after 9:00 a.m. The expression u(8) means the value of the inventory function 8 hours after 9:00 a.m. Since 8 hours after 9:00 a.m. is 5:00 p.
m., the time must be 5:00 p.m., not 10:00 a.m. Now evaluate the function: u(8)=0.21(8)
3
#4.09(8)
2
+18.89(8)+69.5. This gives approximately 66.38, which rounds to about 66 units. Therefore, the correct interpretation is that there are approximately 66 units in the inventory at 5:00 p.m.

NEW QUESTION # 32
The exponential function
f(t)=2900(1.13)
t
represents the size of a bacteria population, where t is the time in hours.
How many bacteria are in the population when t=18?

A. 26,170
B. 20,495
C. 29,573
D. 23,160

Answer: A

Explanation:
This is an exponential growth problem because the base 1.13 is greater than 1. The function starts with 2,900 bacteria and grows by a factor of 1.13 each hour, which represents a 13% hourly increase. To find the population after 18 hours, substitute t=18 into the function: f(18)=2900(1.13)
18
Evaluating the exponent and multiplying by 2900 gives approximately 26,170.38. Since bacteria are counted as whole organisms, this rounds to about 26,170 bacteria. The other answer choices do not match the exponential model at t=18. Therefore, the correct answer is C.

NEW QUESTION # 33
The graphed function v(t) represents the number of vehicles, v, stopped at a toll booth t hours after 6:00 a.m.
The coordinates of points A and B are (1,49.4) and (4,45.8), respectively.

What is the average rate of change of the number of vehicles from point A to point B?

A. -1.2
B. -3.6
C. 47.6
D. 19.04

Answer: A

Explanation:
The average rate of change measures how much the output changes per one unit of input. Here, the input is time in hours after 6:00 a.m., and the output is the number of vehicles stopped at the toll booth. The two given points are A=(1,49.4) and B=(4,45.8). Use the average rate of change formula: (y
2
#y
1
)/(x
2
#x
1
). Substituting the coordinates gives (45.8#49.4)/(4#1). The numerator is #3.6, and the denominator is 3, so the average rate of change is #1.2. The negative sign means the number of vehicles decreased over the interval.

NEW QUESTION # 34
......

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