let A = first card and let B = second card; then A = 3B and B = A-6; since A = 3B then B = (3B) - 6; solve B = 3 and A = 9
3 and 9
Just divide the first number by the second one.
4200 is ten times 420
In the first expression the value is 91 times as large as it is in the second.
The four in the second number is ten times the four on the left of the first number and one hundred thousand times the four on the right.
In the first one, the 3 is ten times the value of the 3 in the second one.
The first value of 5, .55, is 11 times greater than the second value, .05.
10 times in the first number, 0.1 times in the second.
Just divide the first number by the second one.
4200 is ten times 420
The four to the left is ten times the four to the right.
In the first expression the value is 91 times as large as it is in the second.
The second. And it is 1/100 as large as in the first.
Counting 6's from left to right, the first 6 is 6000 and the second 6 is 600 so the first 6 is 10 times the second. If you count from right to left, the is left as an exercise for the student.
It is 1 in 17. Here's why: There are 52 cards in a deck, 13 different value cards (A, 2, 3, etc.), and two cards are dealt. We're finding the odds of getting any pocket pair, not just one. This means that the first card dealt can be any value. The second card has to repeat the value of the first card. For example, the first card dealt is a 10, and now the second card has to be a 10. After the first 10 is dealt, there are 51 cards left in the deck, and 3 different 10's. Therefore, 3 over 51 simplifies to 1 over 17.
If a DIE (not dice) is rolled 90 times, the expected value of the sum of the first and second rolls is 7 if you assume that the die is fair. It does not matter how many times you roll the die, as long as it is at least 2.
the value is nothing there not baseball cards.
The four in the second number is ten times the four on the left of the first number and one hundred thousand times the four on the right.