Please help! I've been on this question for more than an hour... You pick a card at random. Without putting the first card back, you pick a second card at random.

What is the probability of picking an even number and then picking an even number?

Simplify your answer and write it as a fraction or whole number.

1/6

Step-by-step explanation:

Out of these four card choices, for the first pick there are two even cards and four cards in total. This means that on the first pick there is a 2/4=1/2 chance that you pick an even card. On the second pick, if you do not replace the card, then there is 1 even card remaining, and 3 cards in total, leaving a probability of 1/3. Multiplying these two probabilities together, you get an overall chance of 1/6. Hope this helps!

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)