Dividing the given polynomial by (x -6) gives quotient Q(x) and remainder 5 then for Q(-6) = 3 , P(-6) = -31and P(6) =5.
As given in the question,
P(x) be the given polynomial
Dividing P(x) by divisor (x-6) we get,
Quotient = Q(x)
Remainder = 5
Relation between polynomial, divisor, quotient and remainder is given by :
P(x) = Q(x)(x-6) + 5 __(1)
Given Q(-6) = 3
Put x =-6 we get,
P(-6) = Q(-6)(-6-6) +5
⇒ P(-6) = 3(-12) +5
⇒ P(-6) =-36 +5
⇒ P(-6) = -31
Now x =6 in (1),
P(6) = Q(6)(6-6) +5
⇒ P(6) = Q(6)(0) +5
⇒ P(6) = 5
Therefore, dividing the given polynomial by (x -6) gives quotient Q(x) and remainder 5 then for Q(-6) = 3 , P(-6) = -31and P(6) =5.
The complete question is:
Dividing the polynomial P(x) by x - 6 yields a quotient Q(x) and a remainder of 5. If Q(-6) = 3, find P(-6) and P(6).
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Answer:
shapes
Step-by-step explanation:
The patterns are shapes
Density = m/v with mass being in grams and volume in mL......
192 grams/250 mL = 0.768 g/mL
Remark
The point value is (-2,5) So we know the two sides. We need the hypotenuse. We should notice that the x value is minus (-2) and value is y value is plus (5). That means we are in quad 2. Be careful how you read that. (-2,5) is a point. It is not a tangent.
Step One
Find the hypotenuse.
a = - 2
b = 5
c = ??
c^2 = a^2 + b^2
c^2 = (-2)^2 + 5^2
c^2 = 4 + 25
c^2 = 29 Take the square root of both sides.
sqrt(c^2) = sqrt(29)
c = sqrt(29)
Step Two
Find the Cosine of the angle.
Cosine(theta) = adjacent / hypotenuse
Cosine(theta) = -2 / sqrt(29) <<<<<<< Answer
Again, watch out for what you are given.
Answer:
Margin of Error = 5.4088 ;
Confidence interval = (30.1 ; 40.9)
Interval estimate are almost the same
Step-by-step explanation:
Given that :
Population standard deviation, σ = 9.3
Sample size, n = 8
Xbar = 35.5
Confidence level = 90%
The confidence interval:
Xbar ± Margin of error
Margin of Error = Zcritical * σ/sqrt(n)
Zcritical at 90% = 1.645
Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088
Confidence interval :
Xbar ± Margin of error
35.5 ± 5.4088
Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1
Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9
(30.1 ; 40.9)
T distribution =. (30.5 ; 40.5)
Normal distribution = (30.1, 40.9)
