The answer is 3/5, or 60%
11 (comedies) + 19 (tragedies) = 30
30/50 = 3/5, or 60%
The question given is incomplete, I googled and got the complete question as below:
You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.
a. What is the probability that your last pot will have the necessary 7 crabs?
b. What is the probability that your last pot will be empty?
Answer:
a. Probability = 0.0083
b. Probability = 0.0907
Step-by-step explanation:
This is Poisson distribution with parameter λ=2.4
a)
The probability that your last pot will have the necessary 7 crabs is calculated below:
P(X=7)= {e-2.4*2.47/7!} = 0.0083
b)
The probability that your last pot will be empty is calculated as:
P(X=0)= {e-2.4*2.40/0!} = 0.0907
90 attempts are made and only one third are successful. Multiply the attempts by successes and you get 30 successful shots (90 x 1/3= 30).
The answer is 1/1073741824 or about 9.313 * 10^-10
Answer:
17rx2−23rx−71x+75
Step-by-step explanation:
(17x−23)(xr−4)−(3x+17)
=(17x−23)(xr−4)+−1(3x+17)
=(17x−23)(xr−4)+−1(3x)+(−1)(17)
=(17x−23)(xr−4)+−3x+−17
=(17x)(xr)+(17x)(−4)+(−23)(xr)+(−23)(−4)+−3x+−17
=17rx2+−68x+−23rx+92+−3x+−17
=17rx2+−68x+−23rx+92+−3x+−17
=(17rx2)+(−23rx)+(−68x+−3x)+(92+−17)
=17rx2+−23rx+−71x+75
