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if you spin the spinner 3,200 times, what is the best prediction possib…

Question

if you spin the spinner 3,200 times, what is the best prediction possible for the number of times it will not land on blue?

Explanation:

Step1: Count total and non - blue sections

The spinner is divided into 8 sections? Wait, no, let's count again. Looking at the spinner: blue sections, let's see. Wait, the spinner has, let's count the slices. Let's see: blue, blue, blue, blue, blue, blue, then one orange and one green? Wait, no, let's count the number of sections. Wait, the spinner: let's see, the blue parts: how many? Wait, maybe the spinner is divided into 8 equal sections? Wait, no, looking at the image: the spinner has, let's count the number of slices. Let's see, the blue sections: 6, orange:1, green:1? Wait, no, maybe 8? Wait, no, let's check again. Wait, the spinner: let's count the number of sections. Let's see, the total number of sections: let's count the angles. Wait, maybe the spinner is divided into 8 equal parts? Wait, no, the blue parts: 6, orange:1, green:1. So total sections: 6 + 1+ 1 = 8? Wait, no, 6 blue, 1 orange, 1 green: total 8 sections. Then the number of non - blue sections: orange (1) + green (1)=2? Wait, no, wait, maybe I miscounted. Wait, no, let's look again. Wait, the spinner: the blue sections: let's count the number of blue slices. Let's see, the spinner has 6 blue slices, 1 orange, and 1 green. So total slices: 6 + 1+ 1 = 8. The number of non - blue slices: 1 (orange)+1 (green) = 2? Wait, no, that can't be. Wait, maybe the spinner is divided into 8 equal parts, and the non - blue parts are 2? Wait, no, wait, maybe I made a mistake. Wait, let's re - examine. Wait, the spinner: let's count the number of sections. Let's see, the blue sections: 6, orange:1, green:1. So non - blue sections: 2. Wait, no, that seems wrong. Wait, maybe the spinner is divided into 8 sections, and the number of non - blue sections is 2? Wait, no, maybe the total number of sections is 8, and the number of blue sections is 6, non - blue is 2. Then the probability of not landing on blue is the number of non - blue sections divided by total sections, which is $\frac{2}{8}=\frac{1}{4}$? Wait, no, that can't be. Wait, maybe I miscounted the sections. Wait, let's count again. Wait, the spinner: looking at the image, the blue parts: let's see, the spinner has 6 blue, 1 orange, 1 green. So total 8. So non - blue is 2. Wait, but maybe the total is 8, and non - blue is 2. Then the probability of not landing on blue is $\frac{2}{8}=\frac{1}{4}$? Wait, no, that would mean that the number of non - blue times is $3200\times\frac{1}{4}=800$, but that seems low. Wait, maybe I miscounted the sections. Wait, maybe the spinner is divided into 8 sections, and the number of non - blue sections is 2? Wait, no, maybe the total number of sections is 8, and the number of blue sections is 6, so non - blue is 2. Wait, but let's check again. Wait, maybe the spinner is divided into 8 equal parts, and the non - blue parts are 2. Then the probability of not landing on blue is $\frac{2}{8}=\frac{1}{4}$. But wait, maybe I made a mistake. Wait, no, let's think again. Wait, maybe the spinner has 8 sections, 6 blue, 1 orange, 1 green. So non - blue sections: 2. Then the probability of not landing on blue is $\frac{2}{8}=\frac{1}{4}$? Wait, no, that would be if non - blue is 2. But wait, maybe I miscounted the blue sections. Wait, maybe the blue sections are 6, and non - blue are 2 (orange and green). So total sections: 8. Then the number of times it will not land on blue is total spins times the probability of not landing on blue. Probability of not landing on blue is $\frac{\text{number of non - blue sections}}{\text{total number of sections}}=\frac{2}{8}=\frac{1}{4}$? Wait, no, that can't be.…

Answer:

800