An arrow on a spinner is spun 250 times. Some of the results are shown below. Number shown & 1 & 2 & 3 & 4 & 5 Frequency & 25 & 53 & 62 & & The relative frequency of landing on a 4 is the same as the relativ frequency of landing on a 5 Work out the relative frequency of landing on a 4

## Step 1: Understanding the Problem<br /><br />Given in the problem that the arrow was spun 250 times and some of the frequencies (events) are given for the numbers the spinner landed on. We know that number 4 and 5 have the relative frequency, but we don't know their individual frequencies.<br /><br />## Step2: Deduce the Accumulated Frequency or Outcomes for 4 + 5<br /><br />Since the arrow was spun 250 times, whatever the accumulated outcome for the copper spun on 4 & 5 should sum up with the already known outcomes for 1, 2, and 3 which sum = (25+53+62=140).<br /><br />### \( 250 - 140 = 110 \)<br /><br />## Step3: Divide the Remaining Number <br /><br />As the question implies, spinner number 4 and 5 have the same relative frequency, so we should divide the 110 outcomes by 2 or can we written as \( \frac{110}{2} = 55 \).<br /><br />## Step4: Calculate for the Relative Frequency <br /><br />Relative frequency is the fraction of times a spinner lands on specific numbers (events). In this case, we have total spins as 250 and, for numbers 4 and 5 as same 55 spins. The relative frequency will be in more simple form as \( \frac{spinner_4 or sqpinner_5 spins}{total spinns} = \frac{55}{250} = \frac{11}{50} \)