From the set A={3,root2,2root3,root9,root7},a number is selected at random.find the probability that is a rational number.

A = { 3, √2, 2√3, √9, √7} , but √9 = 3 which is rational,

so you have 2 rationals out of the 5

well ?

To determine the probability of selecting a rational number from the set A, we need to first understand what makes a number rational.

A rational number is one that can be expressed as the ratio of two integers, where the denominator is not zero. In other words, it can be written in the form p/q, where p and q are integers and q is not equal to zero.

Looking at the set A, we can see that there are both rational and irrational numbers present. Irrational numbers are numbers that cannot be expressed as a fraction or a ratio. In this case, the irrational numbers are √2, 2√3, and √7.

To find the probability of selecting a rational number from set A, we need to determine the number of rational numbers in the set and divide it by the total number of numbers in the set.

Let's analyze the set A:

A = {3, √2, 2√3, √9, √7}

The rational numbers in set A are 3 and √9. Both of these numbers can be expressed as the ratio of two integers: 3 can be written as 3/1, and √9 is equal to 3/1 as well.

Therefore, the set of rational numbers in A is {3, √9}, which has a total of 2 numbers.

To find the probability, we divide the number of rational numbers by the total number of numbers in A:

Probability = Number of rational numbers / Total number of numbers

Probability = 2 / 5

Therefore, the probability of selecting a rational number from set A is 2/5 or 0.4 (40%).