please help!

1.) A random digit from 1 to 9 (inclusive) is chosen, with all digits being equally likely. The probability that when it's squared it will end with the digit 1.

2.) A random number between 1 and 20 (inclusive) is chosen. The probability that its square root will not be an integer.

See response to:

http://www.jiskha.com/display.cgi?id=1344678011

and post if you need more explanations or would like to post your attempt.

1.

Note that
21*21=441
49*49=2401
Does that give you a hint?

2.
The question looks for
"The probability that its square root will not be an integer"
so check your answer accordingly.

Question: why is it not 2/9? Since I know 1 squared will be 1, which ends in 1, but what about 9? 9 squared is 81, it ends in 1. So that makes 2/9 = 0.22

Question: why is it not 2/9? Since I know 1 squared will be 1, which ends in 1, but what about 9? 9 squared is 81, it ends in 1. So that makes 2/9 = 0.22

@MathMate

1.) Oh, the mysterious world of squaring digits! Let's break it down, shall we? Out of the 9 digits, only 3 and 7, when squared, end with the digit 1. So the probability is 2 out of 9, or as a fraction, 2/9. Just remember to approach those squared digits with caution, they can be quite mischievous!

2.) Ah, the enchanting realm of numbers and their square roots! To find the probability that the square root is not an integer, we need to determine how many numbers between 1 and 20 (inclusive) have non-integer square roots. Turns out, there are 14 such numbers! So, the probability is 14 out of 20, or as a fraction, 14/20. It seems the square roots like to dance to their own rhythm, not conforming to the neat integer steps.

Sure, I can help you with that!

1.) To find the probability that when a randomly chosen digit from 1 to 9 is squared, it will end with the digit 1, we can analyze the patterns in the squares of these numbers.

First, let's find the square of each individual digit from 1 to 9:
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81

Now, let's look at the last digit of each square. We can observe that only the squares of 1 and 9 end with the digit 1. Therefore, the only possible choices that will result in a square with a last digit of 1 are 1 and 9.

Since there are 9 digits to choose from initially, the probability of choosing a digit that will result in a square with a last digit of 1 is 2 (1 and 9) out of 9 possible choices. Therefore, the probability is 2/9.

2.) To find the probability that a randomly chosen number between 1 and 20 (inclusive) will have a square root that is not an integer, we can consider the perfect squares within this range.

Between 1 and 20, the perfect squares are: 1, 4, 9, 16.

Out of the 20 possible numbers, these 4 numbers have integer square roots, leaving us with 16 numbers that have non-integer square roots.

Therefore, the probability of choosing a number between 1 and 20 (inclusive) that has a square root that is not an integer is 16 out of 20, which can be simplified to 4/5.

thank you for your help!

can you check if it's correct?

1.) 1/9
2.) 4/2= 1/5