Answer: 5.3
Step-by-step explanation: Do 9.8 (Friend swim depth) - 4.5 (dolphin jump height) to get your answer.
Answer:
See explaination
Step-by-step explanation:
Probability is the number of desired outcomes over the total outcomes.
1) You want to find Oranges, so that is your desired outcome:
There are 4 oranges in the bowl.
Now sum up all the fruits to get the total:
9+4+7+3+6 = 29
Therefore you have 4/29
2) Same idea, count the peaches and put it over total:
3/29
3) This time, you still want to use the same idea, but its just your desired outcomes that have increased. So instead of just having one fruit, now you want to add the total of 2 fruits as your desired outcome.
(9+4)/29
= 13/29
4) To find the probability of a fruit other than a plum, you can use complementary counting. Since the maximum probability is one, we can find what we dont want, and the outcome of that subtracted from one must be the results that we want.
So a plum is 6/29
1-6/29 = 23/29
5) Same idea as what is used in problems 3 and 4, just a combination of the 2.
(4+3)/29
1-7/29
= 22/29
Answer:
Step-by-step explanation:
a) 
Substitute limits to get
= 
Thus converges.
b) 10th partial sum =

=
c) Z [infinity] n+1 1 /x ^4 dx ≤ s − sn ≤ Z [infinity] n 1 /x^ 4 dx, (1)
where s is the sum of P[infinity] n=1 1/n4 and sn is the nth partial sum of P[infinity] n=1 1/n4 .
(question is not clear)
Answer:67738373783
Step-by-step explanation: