A snowstorm took place during the weekend. Meteorologists tracked the total amount of snowfall for the storm. For the first 2 hours, it snowed at a rate of 1 inch per hour. The snow stopped for an hour and a half, then resumed at a rate of 1.5 inches per hour for the next 3 hours. Based on this information, what scale would work best for representing the data on the y-axis?(1 point)

Question

Responses

0.5
0.5

2
2

3
3

5

Answer 1

To determine the best scale for representing the data on the y-axis, we first need to calculate the total snowfall during the snowstorm.

1. For the first 2 hours, it snowed at a rate of 1 inch per hour:

   • 2 hours × 1 inch/hour = 2 inches.

2. Then it stopped for 1.5 hours (no snow during this time).

3. The snow resumed at a rate of 1.5 inches per hour for the next 3 hours:

   • 3 hours × 1.5 inches/hour = 4.5 inches.

Now, we can calculate the total snowfall:

• Total snowfall = 2 inches + 0 inches (during the stop) + 4.5 inches = 6.5 inches.

Considering the total snowfall of 6.5 inches, the best scale for the y-axis would be one that can adequately display this amount. Here are the provided options:

• 0.5
• 2
• 3
• 5

Given that the total snowfall is 6.5 inches, none of the provided options could represent the total directly without scales that could accommodate a scale larger than 6. However, if we must choose the largest provided option, 5 inches would be the closest choice, as it gives some representation of the scale, even though it does not cover the total snowfall completely.

So, the best response would be: 5.

Answer 2

A dog groomer charges her clients by the weight of the dog. The following piecewise function represents the total charge, f(x)
, for a dog weighing x pounds.

f(x)=⎧⎩⎨⎪⎪2.5x if 0≤x≤203.5x if 20 <x≤50 5x if x>50


What would be the last point graphed on the first piece of the function, f(x)=2.5x
?

(1 point)
(
,
)
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Answer 3

To determine the last point graphed on the first piece of the piecewise function \( f(x) = 2.5x \), we need to look at the given conditions.

The function \( f(x) = 2.5x \) applies for the range \( 0 \leq x \leq 20 \). The last point of this segment occurs at the endpoint of this range.

To find the last point, we evaluate \( f(x) \) at the upper limit of the range:

1. Find \( f(20) \): \[ f(20) = 2.5 \times 20 = 50 \]

Thus, the last point graphed on the first piece of the function is: \[ (20, 50) \]

Therefore, the answer is: \[ (20, 50) \]

Answer 4

Use the image to answer the question.

A piecewise linear function consisting of five parts is plotted in quadrant 1 of a coordinate plane.

On Friday, Kaitlyn went to the park to work out. She started with a slow jog at a constant rate for five minutes. She then increased her speed to a faster, constant pace for seven minutes. She took a three-minute break, then ran for another 7.5 minutes at a constant pace to finish her two-mile run. She rested for the remainder of the time. During what domain interval will she reach 1.25 miles?

(1 point)
Responses

12≤s≤15
12 less-than-or-equal-to s less-than-or-equal-to 15

22.5≤s≤25
22.5 less-than-or-equal-to s less-than-or-equal-to 25

15≤s≤22.5
15 less-than-or-equal-to s less-than-or-equal-to 22.5

5≤s<12
5 less-than-or-equal-to s less than 12
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Answer 5

To determine the domain interval during which Kaitlyn reaches 1.25 miles during her workout, we first need to analyze her workout:

1. First 5 minutes: Slow jog at a constant speed (Let's denote this speed as \( v_1 \)).
2. Next 7 minutes: Increased speed (denote this speed as \( v_2 \)).
3. 3-minute break: No distance covered.
4. Next 7.5 minutes: Running at a constant speed to finish her 2-mile run (denote this speed as \( v_3 \)).

Since we need to find when she reaches 1.25 miles, if we assume her jog (first segment) took her a certain distance, the intervals will depend on the speeds she was running at each of those intervals.

Without the specifics of the graph or the speeds involved, we can infer from typical piecewise linear function behavior that:

• In the first segment, she covers some distance (let's say it's a fraction of a mile).
• By the time she reaches 12 minutes (5 minutes jog + 7 minutes increased pace), she may have reached 1.25 miles if her speeds are sufficient.

However, once she takes a break, she won't be accumulating any distance, so we need to observe how far she’s run before that break and during the subsequent runs to reach a total of 2 miles.

Given the intervals listed, we want to know during which interval she hits the 1.25-mile mark. Typically, we might find that in the first 12 minutes she could reach this mark, but it would also depend on the transition to the second segment of increases.

Based on the response choices provided, the choice 12 ≤ s ≤ 15 is likely the most accurate interval where she reaches 1.25 miles, considering she starts with a jog and accelerates throughout her workout.

Thus, the answer is: 12 ≤ s ≤ 15.

Answer 6

A scientist is observing a particular species of bacteria growing in a petri dish. The number of bacteria can be represented by the equation n(t)=50(1.2)t
, where n is the number of bacteria after t hours. If the scientist checks the population in 4 hours, what would be an appropriate domain for the problem?

Option #1: 0≤t≤104


Option #2: 0≤n≤104


Option #3: 0≤t≤4


Option #4: 0≤n≤4


(1 point)
Option #
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Answer 7

In this scenario, we are examining the growth of a bacteria population over time, where the number of bacteria \( n(t) \) is given by the equation \( n(t) = 50(1.2)^t \).

Given that the scientist is checking the population after 4 hours, we need to determine an appropriate domain related to the time \( t \).

Analyzing the Options:

1. Option #1: \( 0 \leq t \leq 10^4 \)

   • This domain suggests time could go up to 10,000 hours, which is not relevant for an observation limited to 4 hours.

2. Option #2: \( 0 \leq n \leq 10^4 \)

   • This domain relates to the number of bacteria rather than time. While it could represent a possible range for the number of bacteria, it does not define the appropriate time interval for the observation.

3. Option #3: \( 0 \leq t \leq 4 \)

   • This domain represents the time from the start of the observation (0 hours) up to the point of checking the population (4 hours). This is relevant to the specific context of the problem.

4. Option #4: \( 0 \leq n \leq 4 \)

   • This domain also relates to the number of bacteria, which again is not pertinent to the time interval during which the observation is made.

Conclusion:

The appropriate domain for this problem, focusing on the time \( t \) of the observation, is: Option #3: \( 0 \leq t \leq 4 \).

Answer 8

Use the table to answer the question.

x f(x)
0 32,000
1 29,440
2 27,048.80
3 ?
Sydney bought a new car for $32,000.00. If the car depreciates in value by 8% each year, the situation can be modeled by the equation f(x)=32,000(0.92)x
. Complete the table to find the value of the car to the nearest cent after three years.(1 point)
The value of the car after three years is $
.

Answer 9

To find the value of the car after three years, we can use the depreciation formula given:

\[ f(x) = 32,000(0.92)^x \]

where \( x \) is the number of years.

Given the values for \( x \) and \( f(x) \) in the table:

• For \( x = 0 \): \( f(0) = 32,000 \)
• For \( x = 1 \): \( f(1) = 29,440 \)
• For \( x = 2 \): \( f(2) = 27,048.80 \)

Now, we need to calculate \( f(3) \):

\[ f(3) = 32,000(0.92)^3 \]

Calculating \( (0.92)^3 \):

\[ (0.92)^3 = 0.778688 \]

Now, plug this back into the equation:

\[ f(3) = 32,000 \times 0.778688 \approx 24,979.01 \]

So, the value of the car after three years is approximately $24,979.01.

Thus, the value of the car after three years is $24,979.01.

Answer 10

Use the image to answer the question.

An increasing curve with an arrow on the upper end is plotted on a coordinate plane. The x-axis is labeled time left parenthesis years right parenthesis. The y-axis is labeled amount left parenthesis dollars right parenthesis.

Lincoln’s parents invested $5,000 in a college savings account when he was born. If the account has a growth rate of 12%, the situation can be modeled by the equation a(t)=5,000(1.12)t
, where a is the amount after t years. Which one of the following statements is true about the graph of the function?

Option #1: As t increases, a increases slowly at first and then quickly.

Option $2: As t increases, a increases quickly at first and then slowly.

Option #3: As t increases, a decreases slowly at first and then quickly.

Option #4: As t increases, a decreases quickly at first and then slowly.

(1 point)
Option #
is the true statement.
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Answer 11

To analyze the given function \( a(t) = 5000(1.12)^t \), we need to consider how the value \( a(t) \) changes as time \( t \) increases.

Characteristics of Exponential Growth:

1. Initial Growth: At the beginning (when \( t \) is small), the value of \( a(t) \) increases, but the increase is relatively small due to the exponent being relatively low.

2. Diminishing Returns vs. Compounding: As \( t \) increases, the base of the exponent (1.12) means that the function will grow faster over time due to the compounding effect of the interest. This means the more time passes, the larger \( t \) becomes, and the growth of the amount accelerates.

Evaluating the Options:

• Option #1: As \( t \) increases, \( a \) increases slowly at first and then quickly.

  • This option accurately describes exponential growth, where the growth starts slow and accelerates as time goes on.

• Option #2: As \( t \) increases, \( a \) increases quickly at first and then slowly.

  • This option is incorrect; it does not describe the nature of exponential growth.

• Option #3: As \( t \) increases, \( a \) decreases slowly at first and then quickly.

  • This option is incorrect; our function is growing, not decreasing.

• Option #4: As \( t \) increases, \( a \) decreases quickly at first and then slowly.

  • This option is also incorrect; it misrepresents the function's trend.

Conclusion:

The true statement about the graph of the function is:

Option #1: As \( t \) increases, \( a \) increases slowly at first and then quickly.