WGU Applied-Algebra - WGU Applied Algebra FXO2 PFXP C957

This situation can be modeled using a linear relationship because each sack has the same weight.

We are given:

After 3sacks are removed, the total weight is 116pounds.

After 6sacks are removed, the total weight is 101pounds.

From 3 sacks removed to 6 sacks removed, the number of removed sacks increases by:

6-3=3

During that time, the total weight decreases from 116pounds to 101pounds:

116-101=15

So removing 3 additional sacks decreases the total weight by 15 pounds.

Now divide to find the weight of one sack:

15/3=5

So each sack weighs:

5 " pounds "

Check:

If 3 more sacks are removed and each sack weighs 5 pounds, the total weight should decrease by:

3(5)=15

116-15=101

This matches the given information.

The graph compares the populations of two towns over time.

The horizontal axis represents:

" Time in years "

The vertical axis represents:

" Population in thousands "

The graph shows:

Town A as the solid blue line.

Town B as the dashed blue line.

To determine which town’s population is growing faster, we compare the slopes of the two lines.

In Applied Algebra, the slope of a line represents the rate of change:

" slope " = " change in population " / " change in time "

From the graph:

" Town A grows at about " 0.9 " thousand people per year "

" Town B grows at about " 1.6 " thousand people per year "

Now compare the growth rates:

1.6 > 0.9

So Town B’s population is growing at a faster rate.

The values 8.0and 6.5describe starting populations, not growth rates. Since the question asks about growing at a faster rate, we must compare the slopes.

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 "

The graph is a logistic growth function.

A logistic function often has an S-shape. It begins by increasing slowly, then increases faster and faster, and later levels off as it approaches a maximum carrying capacity.

From year 1 to year 9, the graph is rising and becoming steeper.

That means:

" the wild horse population is increasing "

and the slope is getting larger, so:

" the population is increasing faster and faster "

Therefore, the correct answer is:

â–­ ( " A " )

The function is:

P(t)=24t+185

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In the function:

P(t)=24t+185

the coefficient of tis:

24

So the rate of change is:

24 " fish per month "

Because 24is positive, the fish population is increasing.

The number 185is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=24(0)+185=185

Therefore, the correct interpretation is:

" The number of fish in the lake is increasing at a constant rate of 24 fish per month. "

So the correct answer is:

â–­ ( " B " )

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:

â–­ ( " D " )

The graph shows a decreasing exponential function.

The horizontal axis represents:

" Time in months "

The vertical axis represents:

" New user accounts per day "

The graph starts near 7,000new user accounts per day and decreases over time. However, it does not decrease toward 0. Instead, the graph levels off near a horizontal value around:

y=2,500

This horizontal value is the horizontal asymptote.

A horizontal asymptote shows the long-term value the function approaches. Since the asymptote is above 0, the model predicts that the number of daily new user accounts approaches a positive value, not zero.

So the justified statement is:

" The number of daily new user accounts will not decrease to 0. "

The function is:

D(t)=48t+23

This is a linear function because it has the form:

D(t)=mt+b

where:

48

is the rate of change, and:

23

is the starting distance from the rest stop.

We need to find:

D(1.7)

That means substitute:

t=1.7

into the function.

D(1.7)=48(1.7)+23

First multiply:

48(1.7)=81.6

Now add 23:

D(1.7)=81.6+23

D(1.7)=104.6

So the vehicle is:

104.6 miles

from the rest stop after 1.7 hours.

The graph shows the water volume decreasing over time.

The vertical axis represents:

" Water volume in gallons "

The horizontal axis represents:

" Time in minutes "

The graph is an exponential decay curve. It decreases quickly at first, then levels off.

From the graph, the curve approaches a horizontal value near:

y=1,000

This means the horizontal asymptote is approximately:

y=1,000

Since the graph approaches 1,000 gallons, it will not decrease to 500 gallons according to this model.

The water volume does not change by a constant rate because exponential functions do not have constant additive change. Also, the graph decreases fastest at the beginning, not later.