Question

a spinner contains 10 equal - sized sections. the sections are colored red, green, black, or orange. the spinner is spun 50 times. the spinner landed on: red 19 times, green 11 times, black 4 times, orange 16 times. based on these results, how many sections are most likely each color on the spinner? enter the answers in the boxes. then there is a table with columns color and number of spinner sections and rows for red, green, black, orange with empty boxes to fill.

Explanation:

Step1: Find probability ratios

First, we find the ratio of times each color landed to the total spins (50). For red: $\frac{19}{50}$, green: $\frac{11}{50}$, black: $\frac{4}{50}$, orange: $\frac{16}{50}$.

Step2: Multiply by total sections (10)

To find the most likely number of sections, we multiply each ratio by 10 (total sections).

• Red: $10\times\frac{19}{50} = \frac{190}{50}= 3.8\approx4$ (rounded to nearest whole number as sections are whole)
• Green: $10\times\frac{11}{50}=\frac{110}{50} = 2.2\approx2$
• Black: $10\times\frac{4}{50}=\frac{40}{50}=0.8\approx1$
• Orange: $10\times\frac{16}{50}=\frac{160}{50} = 3.2\approx3$

Wait, but let's check if the sum is 10: 4 + 2 + 1 + 3 = 10. Alternatively, maybe we can use the frequency proportion. Let's recalculate more accurately. The total spins are 50, and total sections 10. So each spin corresponds to 10/50 = 0.2 sections per spin. So:

• Red: 19 * 0.2 = 3.8 ≈ 4
• Green: 11 * 0.2 = 2.2 ≈ 2
• Black: 4 * 0.2 = 0.8 ≈ 1
• Orange: 16 * 0.2 = 3.2 ≈ 3

But let's check another way. The number of sections should be proportional to the frequency. So:

Red: (19/50)*10 = 3.8 → 4 (since we can't have a fraction of a section, we round to nearest whole number, and check total)

Green: (11/50)*10 = 2.2 → 2

Black: (4/50)*10 = 0.8 → 1

Orange: (16/50)*10 = 3.2 → 3

Sum: 4+2+1+3=10, which matches total sections.

Answer:

Red: 4
Green: 2
Black: 1
Orange: 3

(Note: Depending on rounding method, sometimes we can also use the exact proportion. Alternatively, if we consider that the number of sections should be such that the expected frequency matches the observed. But the above is the most reasonable approximation.)