The A-frame cabin shown below is 35 feet wide. The roof of
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Given Information:
Mean weekly salary = μ = $490
Standard deviation of weekly salary = σ = $45
Required Information:
P(X > $525) = ?
Answer:
P(X > $525) = 21.77%
Step-by-step explanation:
We want to find out the probability that a randomly selected teacher earns more than $525 a week.
The z-score corresponding to 0.78 from the z-table is 0.7823
Therefore, there is 21.77% probability that a randomly selected teacher earns more than $525 a week.
How to use z-table?
Step 1:
In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 0.7, 2.2, 1.5 etc.)
Step 2:
Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.78 then go for 0.08 column)
Step 3:
Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.
We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.