Answer:
The question is incomplete, but the step-by-step procedures are given to solve the question.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
That is z with a pvalue of 
, so Z = 2.575.
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M.
The upper end of the interval is the sample mean added to M.
The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).
Using remainder theorem, we get:
Substitute t = 5 into the equation:
Thus, we get a remainder of 93 or (A)
Here, DH = HF
x+3 = 3y
x = 3y-3 ----(I)
GH = HE
4x-5 = 2y+3
4x = 2y+8
Substitute value of x,
4(3y-3) = 2y+8
12y-12 = 2y+8
12y-2y = 12+8
10y = 20
y = 2
Now, substitute it in equation 1,
x = 3(2)-3
x = 6-3 = 3
So, your final answer is x=3 & y=2
Answer:
Step-by-step explanation:
First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items
F(x) = x^3-3x^2+0x+0
We have to divide this by x-2
Leading terms in the dividend and divisor are
x^3 and x
Hence quotient I term would be x^3/x=x^2
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
Multiply x-2 by x square and write below the term and subtract
We get
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
---------------
-x^2+0x
Again take the leading terms and find quotient is –x
x-2) x^3-3x^2+0x+0(x^2-x
x^3-2x^2
---------------
-x^2+0x
-x^2-2x
Subtract to get 2x +0 as remainder.
x-2) x^3-3x^2+0x+0(x^2-x-2
x^3-2x^2
---------------
-x^2+0x
-x^2+2x
-------------
-2x-0
-2x+4
------------------
-4
Thus remainder is -4 and quotient is x^2-x-2
