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Applied-Algebra考題, Applied-Algebra考古題分享, Applied-Algebra新版題庫上線, Applied-Algebra真題, Applied-Algebra認證

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WGU Applied-Algebra Exam Syllabus Topics:

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

>> Applied-Algebra考題 <<

Applied-Algebra考古題分享，Applied-Algebra新版題庫上線

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最新的 Courses and Certificates Applied-Algebra 免費考試真題 (Q30-Q35):

問題 #30
The given function represents the price of a commodity, p, in dollars, based on the number of months, m, since the beginning of 2020.
p(m)=5m+5
What is the average rate of change of the price over the interval m=1to m=10?

  • A. 0
  • B. 1
  • C. 5.91
  • D. 32.5

答案：B

解題說明：
The function is:
p(m)=5m+5
This is a linear function in the form:
p(m)=mx+b
For a linear function, the average rate of change over any interval is the same as the slope.
The slope is the coefficient of m:
5
So the price increases at a constant rate of:
$5 " per month "
We can also verify using the average rate of change formula:
(p(10)-p(1))/(10-1)
Find p(10):
p(10)=5(10)+5=55
Find p(1):
p(1)=5(1)+5=10
Now calculate:
(55-10)/(10-1)
=45/9
=5


問題 #31
A music student attempted to learn a difficult new song. Each day, the student would play the song and be scored on their performance. The graph shows the results.

What was the average daily change in performance from day 16 to day 20?

  • A. 1.38
  • B. 43.86
  • C. 6.88
  • D. 27.5

答案：C

解題說明：
The average daily change is the average rate of change over the interval from day 16 to day 20. From the graph, the performance score at day 16 is approximately 30.1, and the score at day 20 is approximately 57.6.
Use the formula
x
2
#x
1
y
2
#y
1
Substituting gives
20#16
57.6#30.1
The numerator is 27.5, and the denominator is 4, so the average daily change is 27.5/4=6.875, which rounds to 6.88. This means the performance score increased by an average of about 6.88 points per day over the interval. Therefore, the correct answer is B.


問題 #32
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. Exponential
  • B. Linear
  • C. Polynomial
  • D. Logistic

答案：A

解題說明：
The scatterplot shows the number of monthly train crossings decreasing over time. The decrease is not constant from one year to the next, so a linear model is not the best choice. Instead, the data decreases quickly at first and then begins to level off, which is characteristic of exponential decay. Exponential models are appropriate when a quantity changes by a repeated factor or percentage over equal time intervals. A logistic model would usually show an S-shaped pattern with a carrying capacity, and a polynomial model would be more appropriate for turning points or more complex curvature. Since the points show a smooth decreasing curve that flattens over time, an exponential function best models the data. Therefore, the correct answer is B.


問題 #33
The value of an investment portfolio after x years is modeled by the function g(x)=3100(1.059) x
.
What is the value of the portfolio when x=10?

  • A. $5,193.08
  • B. $5,499.48
  • C. $6,167.56
  • D. $5,823.94

答案：B

解題說明：
This problem uses an exponential growth model. The initial portfolio value is 3100, and the growth factor is
1.059. Since the growth factor is greater than 1, the portfolio increases over time. To find the value after 10 years, substitute x=10 into the function: g(10)=3100(1.059)
10
First evaluate the exponent, then multiply by 3100. This gives g(10)#5499.4758. Rounded to the nearest cent, the value is $5,499.48. The answer must be written as money because the function models the value of an investment portfolio. Among the choices, only option B matches the correctly evaluated exponential expression. Therefore, the correct answer is B.


問題 #34
The function f(n) represents the relationship between the distances traveled by two vehicles, where n is the distance traveled by vehicle A and f is the distance traveled by vehicle B. The distance traveled by vehicle B is 17 more than the distance traveled by vehicle A.
Which function represents this situation?

  • A. f(n)=n+17
  • B. f(n)=17n
  • C. f(n)=n#17
  • D. f(n)=n/17

答案：A


問題 #35
......

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