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marishachu [46]
3 years ago
10

In a game of chance, players spin the pointer of a spinner with eight equal-sized sections. A spinner that is divided into 8 equ

al sections labeled 1, 2, 1, 3, 1, 4, 1, and 5. What is the probability that the pointer will land in an odd-numbered section? A. 12 B. 58 C. 34 D. 78
Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0

Answer:

6/8

Step-by-step explanation:

probability= required outcome / all possible outcomes

probability = 6/8

Elza [17]3 years ago
8 0

Answer:

D is the closest, but look at the work below.

Step-by-step explanation:

First let's write what we know.

We have 8 equal sections, they are:

                            1

                     5          2

                 1                     1

                     4          3

                            1

We need to find the probability the pointer will land on an odd-numbered section.

Now we find how many odd numbered sections there are.

                           <em> </em><u><em>1</em></u>

                     <u><em>5</em></u>          2

                 <u><em>1</em></u><u><em> </em></u>                    <em><u>1</u></em>

                     4          <u><em>3</em></u>

                            <u><em>1</em></u>

There are 6 odd numbered sections.

So you have a 6 in 8 chance of spinning an odd number or 6/8.

6/8 can be reduced to 3/4 or .75 05 75%

This doesn't match any of your answers you gave, so double check the question or the answers. D is the closest, but this is a very clear solution and I am wondering if there is an oops somewhere.

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