You reach into a bag of coins and withdrew two coins. what is the probability you withdrew a nickel and then a dime if the bag held ten pennies, six nickels, and five dimes? a. 5/14 b. 30/441 c. 1/14 d. 15/441

1/14

Step-by-step explanation:

The total number of coins in the bag is 10+6+5, or 21.

You reach into the bag and take ONE coin.  The chances of that being a nickel is 6/21, or 2/7, since nickels compose 6 of the 21 coins in the bag.

You don't replace the nickel.

Now you have 20 coins in the bag, including 5 nickels.

The chances of your next draw being a dime is 5/20, or 1/4, since there are 5 dimes in the bag.

The joint probability of drawing a nickel and then a dime is then the product of these two probabilities:

(2/7)(1/4) = 2/28 = 1/14

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.