Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

\begin{tabular}{|c|c|}
\hline
\multicolumn{2}{|c|}{ Hopping } \\
\hline
Number of hops & \begin{tabular}{c}
Distance traveled \\
(f)
\end{tabular} \\
\hline
20 & 30 \\
\hline
50 & 75 \\
\hline
80 & 120 \\
\hline
150 & [tex]$?$[/tex] \\
\hline
\end{tabular}

Which statement correctly explains how to find the distance traveled after 150 hops?

A. Subtract [tex]$120 - 75$[/tex] to get 45, then add that number to 120.

B. Add [tex]$30 + 75 + 120$[/tex].

C. Find a number that when multiplied by 50 (number of hops in the 2nd row) will give a product of 150 (number of hops in the 4th row), then multiply 75 by that number.

Sagot :

To determine the distance traveled after 150 hops, we need to analyze the relationship between the number of hops and the distance traveled. Given the data points:

- 20 hops correspond to 30 units of distance.
- 50 hops correspond to 75 units of distance.
- 80 hops correspond to 120 units of distance.

We can see the relationship between the number of hops and the distance traveled appears to be linear. Let's find the slope (rate of change) between the given data points first:

1. Calculate the slope between the first two data points (20 hops, 30 units) and (50 hops, 75 units):
[tex]\[ \text{Slope}_1 = \frac{75 - 30}{50 - 20} = \frac{45}{30} = 1.5 \][/tex]

2. Calculate the slope between the next two data points (50 hops, 75 units) and (80 hops, 120 units):
[tex]\[ \text{Slope}_2 = \frac{120 - 75}{80 - 50} = \frac{45}{30} = 1.5 \][/tex]

Both slopes are equal, indicating a consistent linear relationship. Therefore, the average slope (rate of change) between hops and distance is:
[tex]\[ \text{Average slope} = \frac{\text{Slope}_1 + \text{Slope}_2}{2} = \frac{1.5 + 1.5}{2} = 1.5 \][/tex]

In a linear equation of the form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept, rearranging one of the known points can help us find [tex]\(b\)[/tex]. Using the point (20, 30):

[tex]\[ 30 = 1.5 \times 20 + b \implies 30 = 30 + b \implies b = 0 \][/tex]

Therefore, the linear equation relating hops to distance is:
[tex]\[ \text{Distance} = 1.5 \times \text{Number of hops} \][/tex]

To find the distance for 150 hops, substitute [tex]\( \text{Number of hops} = 150 \)[/tex] into the equation:

[tex]\[ \text{Distance for 150 hops} = 1.5 \times 150 = 225 \][/tex]

Thus, the distance traveled after 150 hops is 225 units. The correct answer is:

[tex]\[ \text{The distance traveled after 150 hops is 225 units.} \][/tex]