Answer:
Step-by-step explanation:
Let X be length of life of the participants in the plan.
Given that X is N(68,3.5)
We convert this to standard normal score z using
a) proportion of the plan recipients that would receive payments beyond age 75=
b) proportion of the plan recipients die before they reach the standard retirement age of 65=
c) x for 86% ceased
Answer: both the left and right sides go to +∞
<u>Step-by-step explanation:</u>
End behavior can be determined by two things:
- Sign of the leading coefficient
- Degree of the function
<u>Sign of leading coefficient</u>:
positive: right side goes to +∞
negative: right side goes to -∞
⇒ Leading coefficient of this function is 3 so the right side goes to +∞
<u>Degree (exponent of leading coefficient)</u>:
even: both the left and right sides point in the SAME direction
odd: the left and right sides point in OPPOSITE directions
⇒ Degree of this function is 4 so the left side will point in the same direction as the right side.
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
