Use the spinner below to answer the question. Assume that it is equally probable that the pointer will land on any one of the five numbered spaces. If the pointer lands on a​ borderline, spin again. A spinner consists of a pointer on a circle divided into five equally-sized sections by five line segments that run from the circle's center to its edges. In clockwise order, the sectors are labeled "1," "2," "3," "4," and "5." Find the probability that the arrow will land on an odd number.

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Answer:

The probability that the arrow will land on an odd number is 0.60.

Step-by-step explanation:

We are given that a spinner consists of a pointer on a circle divided into five equally-sized sections by five line segments that run from the circle's center to its edges.

In clockwise order, the sectors are labeled "1," "2," "3," "4," and "5."

As we know that the probability of any event is given by;

        Probability, P(E) = [tex]\frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}[/tex]

Now, the probability that the arrow will land on an odd number is given by;

Total number of sectors = 5

Odd number of sectors = 1, 3, 5 = 3

So, the required probability = [tex]\frac{3}{5}[/tex] = 0.60

Hence, the probability that the arrow will land on an odd number is 0.60.

Answer:

Step-by-step explanation:

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