A group of people were asked how much time they spent exercising yesterday. Their responses are shown in the table below.

What fraction of these people spent less than 20 minutes exercising yesterday? Give your answer in its simplest form.

\begin{tabular}{|c|c|}
\hline Time, [tex]$t$[/tex] (minutes) & Frequency \\
\hline [tex]$0 \leq t\ \textless \ 10$[/tex] & 3 \\
\hline [tex]$10 \leq t\ \textless \ 20$[/tex] & 22 \\
\hline [tex]$20 \leq t\ \textless \ 30$[/tex] & 6 \\
\hline [tex]$30 \leq t\ \textless \ 40$[/tex] & 3 \\
\hline [tex]$40 \leq t\ \textless \ 50$[/tex] & 1 \\
\hline
\end{tabular}



Answer :

Sure, let's solve this step-by-step.

1. Identify the total number of people surveyed:

To find the total number of people surveyed, we add up the frequencies of all the intervals listed:
[tex]\[ 3 + 22 + 6 + 3 + 1 = 35 \][/tex]
So, the total number of people surveyed is 35.

2. Identify the number of people who spent less than 20 minutes exercising:

We look at the intervals that are less than 20 minutes. These intervals are [tex]\(0 \leq t < 10\)[/tex] and [tex]\(10 \leq t < 20\)[/tex].
- For the interval [tex]\(0 \leq t < 10\)[/tex], the frequency is 3.
- For the interval [tex]\(10 \leq t < 20\)[/tex], the frequency is 22.

Adding these frequencies together gives us:
[tex]\[ 3 + 22 = 25 \][/tex]
So, 25 people spent less than 20 minutes exercising.

3. Determine the fraction of people who spent less than 20 minutes exercising:

The fraction of people who spent less than 20 minutes exercising is the number of people who spent less than 20 minutes divided by the total number of people surveyed:
[tex]\[ \frac{25}{35} \][/tex]

4. Simplify the fraction:

To simplify the fraction [tex]\(\frac{25}{35}\)[/tex], we divide both the numerator and the denominator by their greatest common divisor (GCD). The GCD of 25 and 35 is 5.
[tex]\[ \frac{25 \div 5}{35 \div 5} = \frac{5}{7} \][/tex]

Hence, the fraction of people who spent less than 20 minutes exercising yesterday is [tex]\(\frac{5}{7}\)[/tex] in its simplest form.