Answer:
0.9905,0.9837,0.8066
Step-by-step explanation:
Given that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).
Each shaft is independent of the other and probability for non conforming is the same for each trial.
Hence X the among these that are nonconforming and can be reworked. is Bin with n =200 and p = 0.10
a) 
b) 
c) 
Answer:
The base (b) has to be positive and different of 1. The logarithm is the inverse of exponential, so:
logb(a) = x ⇒ a = bˣ
So, for b = 0 ⇒ 0ˣ = a
And there is impossible, "a" only could be 0.
For b = 1 ⇒ 1ˣ = a
And the same thing would happen, the logarithming would be to be 1, and the function will be extremally restricted.
For b<0, then the expression a = bˣ will be also restricted, and will not represent all values of a.
So, 0<b<1 and b >1.
Answer:
question 3: 1/3 question 4: 5/7 question 5: 2/3
Answer:
Step-by-step explanation:
Let 
Subbing in:

a = 9, b = -2, c = -7
The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

Group together in groups of 2:

Now factor out what's common within each set of parenthesis:

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

Remember that 
so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring
gives us that x = 1 and -1. The other set is a bit more tricky. If
then
and

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:
±
which will simplify down to
±
Those are the 4 solutions to the quartic equation.