Applied Algebra FX01 Exam (C957)
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Free Applied Algebra FX01 Exam (C957) Questions
The rate that a rumor spreads through a middle school can be modeled with a logistic function. The table below was produced using a logistic function, where R is the number of people who have heard the rumor after d days of it being introduced to the student body.
How should the average rate of change from day 1 to day 5 be interpreted?
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The rumor is spreading at a decreasing rate of 365 people per day.
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The rumor is spreading at an increasing rate of 210 people per day.
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The rumor is spreading at an increasing rate of 365 people per day.
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The rumor is spreading at a decreasing rate of 210 people per day.
Explanation
Solution:
To calculate the average rate of change, we use the formula:
=
= 364.75
Interpretation:
The logistic function represents a rumor spreading at decreasing rates over time as the number of people who have heard the rumor approaches the limit (asymptote). Thus, the rumor spreads at a decreasing rate of 365 people per day from day 1 to day 5.
Correct Answer:
The rumor is spreading at a decreasing rate of 365 people per day.
Why the Other Options Are Incorrect:
"The rumor is spreading at an increasing rate of 210 people per day": The average rate of change is 365, not 210, and the spread of the rumor slows down (decreasing rate), not increases.
"The rumor is spreading at an increasing rate of 365 people per day": While the rate is 365, the logistic function shows decreasing rates over time, not increasing.
"The rumor is spreading at a decreasing rate of 210 people per day": The value 210 is incorrect; the average rate of change is 365.
On a certain day, the electric power in thousands of megawatts used by a major city can be approximated by the following graph below, where f is the number of hours since midnight for 0 ≤ t ≤ 24.
How should the average rate of change between t = 16 and t = 22 be interpreted?
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Between these times, about 500 more megawatts of electricity are used every hour.
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Between these times, about 2,000 fewer megawatts of electricity are used every hour.
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Between these times, about 2,000 more megawatts of electricity are used every hour.
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Between these times, about 500 fewer megawatts of electricity are used every hour.
Explanation
Solution:
From the graph:
At t = 16, P(16) ≈ 21.2 thousand megawatts.
At t = 22, P(22) ≈ 18 thousand megawatts.
Substituting these values into the formula:
= - 0.5333 thousand megawatts.
≈ - 500 megawatts
Correct Answer:
Between these times, about 500 fewer megawatts of electricity are used every hour.
Why other options are wrong:
"Between these times, about 500 more megawatts of electricity are used every hour." This is incorrect because the rate of change is negative, indicating a decrease, not an increase, in electricity usage.
"Between these times, about 2,000 fewer megawatts of electricity are used every hour." This is incorrect because the calculated decrease is approximately 500 megawatts per hour, not 2,000 megawatts.
"Between these times, about 2,000 more megawatts of electricity are used every hour." This is incorrect because the electricity usage decreases (negative rate), so there is no increase in usage.
A machine depreciates in value according to the function: V(t) = 12,000−800t where V(t) represents the value of the machine (in dollars) after t years. After how many years will the machine's value be $4,000?
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5 years
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7 years
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10 years
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15 years
Explanation
Correct Answer:
10 years
Solution:
Substitute V(t) = 4,000 into the equation:
4,000 = 12,000 − 800t
800t = 12,000−4,000
800 t = 8,000
t = 8,000/800
= 10
The machine's value will reach \(4,000 after 10 years. Why Other Options Are Wrong: "5 years" Incorrect because the value at 5 years would be 12,000 − 800(5) = 8,000 12,000 – 800(5) = 8,000 12,000 − 800(5) = 8,000, not\)4,000.
"7 years" Incorrect because the value at 7 years would be
12,000 − 800(7) = 6,400
12,000 - 800(7) = 6,400
12,000−800(7) = 6,400.
"15 years" Incorrect because after 15 years, the value would drop below $4,000.
The adoption rate of a new mobile app in a city can be modeled with a logistic function. The table below shows data from this model, where A is the number of users who have downloaded the app after w weeks since its launch.
How should the average rate of change from week 1 to week 5 be interpreted?
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The app adoption is spreading at a decreasing rate of 234 thousand users per week.
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The app adoption is spreading at an increasing rate of 234 thousand users per week.
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The app adoption is spreading at a decreasing rate of 185 thousand users per week.
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The app adoption is spreading at an increasing rate of 185 thousand users per week.
Explanation
Solution:
To calculate the average rate of change, we use the formula:
= 233.75
Interpretation:
The logistic function demonstrates that the app adoption rate slows down as it approaches its maximum potential user base (asymptote), indicating a decreasing rate over time. Therefore, the app adoption is spreading at a decreasing rate of 234 thousand users per week from week 1 to week 5.
Correct Answer:
The app adoption is spreading at a decreasing rate of 234 thousand users per week.
Why the Other Options Are Incorrect:
"The app adoption is spreading at an increasing rate of 234 thousand users per week": While the rate is correct at 234, the logistic function shows decreasing rates over time, not increasing.
"The app adoption is spreading at a decreasing rate of 185 thousand users per week": The value 185 is incorrect; the actual average rate of change is 234.
"The app adoption is spreading at an increasing rate of 185 thousand users per week": Both the rate (185) and the description of increasing are incorrect.
A technology firm develops the following computer model to help a program that controls a solar array. The y-axis shows the hours of daylight, and the x-axis shows the day of the year, with the solstices and equinoxes being labeled (approximately December 21, March 21, June 21, and September 21).
Estimate and interpret the average rate of change between December 21 and June 21, assuming there are 182 days of the year between them.
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The average rate of change is approximately 0.03 hours per day; this means every day from December 21 to June 21 has about 0.03 more hours of daylight than the day before, on average.
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The average rate of change is approximately 0.5 hours per day; this means every day from December 21 to June 21 has about 0.5 more hours of daylight than the day before, on average.
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The average rate of change is approximately -0.5 hours per day; this means every day from December 21 to June 21 has about 0.5 fewer hours of daylight than the day before on average.
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The average rate of change is approximately -0.03 hours per day; this means every day from December 21 to June 21 has about 0.03 fewer hours of daylight than the day before, on average.
Explanation
Correct answer:
The average rate of change is approximately 0.03 hours per day; this means every day from December 21 to June 21 has about 0.03 more hours of daylight than the day before, on average.
Solution:
Since there are 182 days between June 21 - December 2,
= 0.021978
≈ 0.03
Why other are wrong:
“The average rate of change is approximately 0.5 hours per day; this means every day from December 21 to June 21 has about 0.5 more hours of daylight than the day before, on average.” This option suggests a much larger increase in daylight (0.5 hours per day), which is too high. The correct rate of change is around 0.03 hours per day, not 0.5 hours.
“The average rate of change is approximately -0.5 hours per day; this means every day from December 21 to June 21 has about 0.5 fewer hours of daylight than the day before on average.” This option suggests a decrease in daylight, but from December 21 to June 21, the daylight hours increase (as the days get longer leading up to the summer solstice). Therefore, the rate of change should be positive, not negative.
“The average rate of change is approximately -0.03 hours per day; this means every day from December 21 to June 21 has about 0.03 fewer hours of daylight than the day before, on average.” This option also suggests a decrease in daylight hours, which is incorrect for this time period, as the daylight hours increase towards the summer solstice.
The revenue for a startup over the first six years was modeled using a revenue function R. For instance, R(1)=30,000 indicates the company earned $30,000 in its first year. The revenue function values for the remaining years are:
R(2) = 52,000
R(3)=50,000
R(4)=51,000
R(5) = 80,000
R(6) = 53,000
Which of the following best describes the trend in the revenue values?
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Revenue consistently decreased after year five.
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Revenue steadily increased over the years without any drops.
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After year one, revenue remained close to $50,000, with a noticeable peak in year five.
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Revenue increased every year, reaching its highest point in year five.
Explanation
Correct Answer:
After year one, revenue remained close to \(50,000, with a noticeable peak in year five. Explanation: Analyzing the data: R(1)=30,000: Revenue starts at\)30,000 in year one.
R(2)=52,000: Revenue rises significantly in year two.
R(3)=50,000: Revenue slightly drops but stays near \(50,000. R(4)=51,000: Revenue stabilizes around\)50,000.
R(5)=80,000: Revenue spikes in year five.
R(6)=53,000: Revenue decreases but remains close to \(50,000. Why the other options are incorrect: "Revenue consistently decreased after year five": Revenue in year six was\)53,000, higher than the starting revenue in year one.
"Revenue steadily increased over the years without any drops": Revenue dropped between years two and three, and again after year five.
"Revenue increased every year, reaching its highest point in year five": Revenue did not increase every year—it decreased in years three and six.
The revenue function R(x)=5x3− 30x2 + 100x represents the total revenue (in millions of dollars) generated from selling x units of a product (in thousands). How can the value R(3) be interpreted?
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Selling 3 units generates $315 million in revenue.
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Selling 3,000 units generates $615 million in revenue.
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Selling 3 units generates $85 million in revenue.
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Selling 3,000 units generates $85 million in revenue.
Explanation
Solution:
To interpret R(3), substitute x=3 into the revenue function:
R(x) = 5(3)3− 30(3)2 + 100(3)
R(3) = 5(3)3− 30(3)2 + 100(3)
R(3) = 5(27)−30(9)+100(3)
R(3) = 5(27) - 30(9) + 100(3)
R(3) = 135 - 270 + 300
R(3)=165
Thus, R(3) =165. Since x represents the number of products sold in thousands, selling 3 (in thousands) means selling 3,000 products. The revenue is measured in millions of dollars, so R(3)=165 corresponds to total revenue of \(165 million. Correct Answer: Selling 3,000 units generates\)165 million in revenue.
Why the Other Options Are Incorrect:
"Selling 3 units generates \(315 million in revenue": The function uses x in thousands, so x = 3 refers to 3,000 units, not 3 units. Additionally, the calculated revenue is\)165 million, not \(315 million. "Selling 3 units generates\)85 million in revenue": This incorrectly assumes x = 3 represents 3 units instead of 3,000. Also, the revenue is $165 million, not $85 million.
"Selling 3,000 units generates \(85 million in revenue": The revenue for selling 3,000 units is\)165 million, not $85 million.
The following table shows the revenue and gross profit margin percentage for last year. The first column is the quarter, the second column is the revenue for that quarter, and the final column is the gross profit margin percentage.
Which conclusion can be made from this data?
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Revenues were higher in the last half of the year.
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Revenues were higher in the first half of the year.
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Profits were higher in the last half of the year.
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Profits were higher in the first half of the year.
Explanation
Solution:
First half revenue: \(33,300 +\)34,100 = \(67,400 Last half revenue:\)31,950 + \(35,450 =\)67,400
The revenues are equal in both halves, so this is false
Since we proved the revenues are equal in both halves, this is also false
First half profit margins: 13.13% and 17.67% Last half profit margins: 28.67% and 19% The highest margins were in Q3, and the average margin is higher in the last half, so this is true! Since we proved profits were higher in the last half, this is false
The correct answer:
Profits were higher in the last half of the year.
Why other options are wrong:
“Revenues were higher in the last half of the year.” First half revenue: \(33,300 +\)34,100 = $67,400 Last half revenue: $31,950 + \(35,450 =\)67,400 The revenues are equal in both halves, so this is false
“Revenues were higher in the first half of the year.” Since we proved the revenues are equal in both halves, this is also false
“Profits were higher in the first half of the year” Since we proved profits were higher in the last half, this is false
The population of a species of albino squirrels in a certain city is represented by the graph below, with representing the time in months and P(t) representing the number of squirrels.
What is the maximum albino squirrel population as time increases, according to this graph?
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The population approaches 40 squirrels.
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The population approaches 250 squirrels.
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The population approaches 300 squirrels.
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The population approaches 350 squirrels.
Explanation
Correct answer:
The population approaches 300 squirrels.
Solution:
The curve reaches an asymptote as the number of squirrels approaches 300.
Why others are wrong:
“The population approaches 40 squirrels.” This number is too low. This number was attained after two months.
“The population approaches 250 squirrels.” This number is not the maximum since the graph keeps on rising.
“The population approaches 350 squirrels.” The population of squirrels may never reach this number since the curve has an asymptote at 300 squirrels.
The scatterplot shows data on the number of annual bear sightings in a region over time. The results are shown in the graph. A regression function is graphed with r2 = 0.42. The predicted number of annual bear sightings after 19.5 years is 62.2. 
Is this prediction appropriate?
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Yes. The r2 value indicates a moderate fit, and x = 19.5 is within 25% of the range of the maximum value.
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No. The r2 value indicates a moderate fit, but x = 19.5 is more than 25% of the range beyond the maximum value.
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Yes. The r2 value indicates a strong fit, and x = 19.5 is within 50% of the range of the maximum value.
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No. The r2 value indicates a strong fit, but x = 19.5 is more than 50% of the range beyond the maximum value.
Explanation
Explanation
In this case, the maximum time is 16 years and the minimum is 6 years. The predicted value occurs at x = 19.5, which is 3.5 years beyond the maximum value of 16 years. The difference between 16 and 19.5 is 3.5, and the total range from the minimum to the maximum is 16 - 6 = 10. The predicted time of 19.5 years is more than 35% beyond the maximum value. This means that the prediction is made beyond an acceptable range, considering the moderate r2 value of 0.42.
Correct answer
No. The r2 value indicates a moderate fit, but x = 19.5 is more than 25% of the range beyond the maximum value.
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