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3 years ago

Based on the polynomial remainder theorem, what is the value of the function when x = 4?

Mathematics

1 answer:

ipn [44]3 years ago

4 0

Answer:

124

Step-by-step explanation:

When x=4, substitute 4 into the function to give f(4).

$f(4) = (4)^4 - 2(4)^3 + 5(4)^2 -20(4) -4\\f(4) = 256 - 2(64) + 5(16) - 80 -4\\f(4) = 256 -128 + 80 - 80 -4\\f(4) = 128+80-80-4\\f(4) = 208-80-4\\f(4) = 128-4\\f(4) = 124$

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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

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Mashcka [7]

Answer:

Step-by-step explanation:

(x+5)(x+2)=x^2+7x+10

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3 years ago

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DanielleElmas [232]

Answer:

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(b) No

Step-by-step explanation:

Given

See attachment

Required

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Using: $F + V - E = 2$

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Answer:

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