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

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.

                            1

                     5          2

                 1                    1

                     4          3

                            1

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.

english, cause i can’t you

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,