Applied-Algebra Latest Test Sample - Applied-Algebra Authorized Exam Dumps

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WGU Applied-Algebra Exam Syllabus Topics:

SectionObjectives
Topic 1: Linear Functions- Interpret inputs and outputs of linear functions
- Interpret rates of change for linear functions
Topic 2: Graphical Depictions- Interpret concavity for situations
- Interpret maximum and minimum for situations
- Interpret asymptotes for situations
- Interpret inputs and outputs for situations
- Interpret rates of change for situations
Topic 3: Logistic Functions- Interpret rates of change for logistic functions
- Interpret inputs and outputs for logistic functions
- Interpret asymptotes for logistic functions
Topic 4: Polynomial Functions- Interpret rates of change for polynomial functions
- Interpret concavity for polynomial functions
- Interpret inputs and outputs for polynomial functions
Topic 5: Functions and Algebra of Functions- Derive conclusions based on data
- Derive conclusions based on notation
- Derive conclusions based on graphs
Topic 6: Validity of Models- Determine fit
- Examine utility
- Determine validity
Topic 7: Exponential Functions- Interpret rates of change for exponential functions
- Interpret inputs and outputs for exponential functions
- Interpret concavity for exponential functions

>> Applied-Algebra Latest Test Sample <<

Hot Applied-Algebra Latest Test Sample | Valid WGU Applied-Algebra: WGU Applied Algebra FXO2 PFXP C957 100% Pass

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WGU Applied Algebra FXO2 PFXP C957 Sample Questions (Q111-Q116):

NEW QUESTION # 111
The number of comments on a social media post is represented by the logistic function f(x), whose graph is shown, where xrepresents the number of days since the post was created and f(x)represents the number of comments on day x.

How does the number of comments change as time progresses from day 1 to day 17?

Answer: A

Explanation:
This graph represents a logistic growth function, which has an S-shaped curve.
Key behavior of logistic functions:
Initial phase (early time) # growth is slow
Middle phase # growth speeds up (increasing rate)
Later phase # growth slows down as it approaches a maximum
Analyze the interval from day 1 to day 17:
This interval is in the early part of the graph.
The curve is increasing and getting steeper over this range.
That means the rate of increase is growing over time.
So, the number of comments is:
" increasing faster and faster "


NEW QUESTION # 112
Consider the function I(m)=41m²#490m+30, which represents the daily income, I, in dollars, of a store based on the number of months, m, since the beginning of the year. What is the difference between I(7) and I(9)?

Answer: B

Explanation:
This is a polynomial function evaluation problem. Since I(m) represents daily income in dollars, the difference between I(7) and I(9) must also be in dollars. First evaluate I(7)=41(49)#490(7)
+30=2009#3430+30=#1391. Then evaluate I(9)=41(81)#490(9)+30=3321#4410+30=#1059. The difference is
|#1059#(#1391)|=332. Because the question asks for the difference between the two outputs, the correct unit is dollars, not months. So the difference between I(7) and I(9) is approximately $332. Therefore, the correct answer is C.


NEW QUESTION # 113
The function P(t)represents the daily profit, in hundreds of dollars, for a museum since opening. The graph of P(t)is shown.

What is the correct interpretation of the maximum value?

Answer: D

Explanation:
The graph shows a curved, downward-opening function. This type of graph is commonly associated with a quadratic polynomial function.
The maximum value of a downward-opening parabola occurs at its highest point, called the vertex.
From the graph, the highest point occurs at approximately:
t=9.5
This means the museum reaches its maximum daily profit approximately:
9.5 " years after opening "
The vertical axis represents daily profit in hundreds of dollars. From the graph, the maximum P(t)-value is approximately:
16.5
Since the profit is measured in hundreds of dollars:
16.5×100=1650
So the maximum daily profit is approximately:
$1,650
Therefore, the correct interpretation is:
" Approximately 9.5 years after opening, a maximum daily profit of approximately " $1,650 " was earned. "


NEW QUESTION # 114
The function a(x)represents the value of an investment account, where xis the number of years since the account was opened. The graph of the function is shown.

What is the total value when x=4?

Answer: B

Explanation:
The function a(x)represents the total value of an investment account.
The horizontal axis represents:
x= " time in years "
The vertical axis represents:
a(x)= " total account value in dollars "
We are asked to find the total value when:
x=4
To answer this from the graph:
Locate x=4on the horizontal axis.
Move vertically upward until reaching the blue graph.
Read the corresponding value on the vertical axis.
From the graph, when x=4, the value is slightly above $1,800, approximately:
a(4)#1,830.59
So the total value of the investment account after 4 years is approximately:
$1,830.59


NEW QUESTION # 115
An investment account accrues interest every year, and the value of the account is given by G(x). The graph of this function is shown.

What represents the value of the account after 4 years?

Answer: D

Explanation:
The graph represents the value of an investment account over time. The horizontal axis gives time in years, and the vertical axis gives the total account value in dollars. To find the value after 4 years, locate x=4 on the horizontal axis and read the corresponding G(x)-value on the graph. The plotted exponential growth curve is slightly above $1,000 at 4 years, closest to $1,052.87. Because the graph shows an increasing exponential trend, values such as $480.52, $584.62, and $711.28 are too low for the account value at x=4. The correct interpretation is that after 4 years, the investment account is worth approximately $1,052.87. Therefore, the correct answer is D.


NEW QUESTION # 116
......

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