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4 years ago

What is the remainder when 4x^3+2x^2-18+38/x-3

Mathematics

1 answer:

Trava [24]4 years ago

7 0

Answer:

The remainder is 2

Step-by-step explanation:

Hope this helps uwu

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lilavasa [31]

It is the answer D, because 9 it’s in all answers

3 0

3 years ago

Mrs. Williams needed to make 4 1/2 of fruit salad for a class party.

ziro4ka [17]

Answer:

strawberry=7/8

Step-by-step explanation:

treat strawberry as x

add on proportions

1 1/8+ 3/4+ 1 3/4 + x = 4 1/2

(9+6+14)/8 + x = 9/2

29/8 +x = 9/2

x= 9/2 - 29/8

x= (36-29)/8

x=7/8

3 0

3 years ago

On a quiz worth 5 points, two students earned a 5, zero students earned a 4, zero students earned a 3, zero students earned a 2,

olchik [2.2K]

The class average would be all the test scores added up and divided by the number of scores.

5+5+1+1+1+1+1+0+0+0+0= 15

15/11=1.37

The class average is 1.37

8 0

4 years ago

Read 2 more answers

The Day before the day before yesterday was two days after the day before my birthday. today is Thursday what day was my birthda

oksian1 [2.3K]

Answer:

Saturday?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a

Nana76 [90]

Answer:

0.00002 = 0.002% probability of actually having the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Having the disease

Probability of having a positive test:

0.05 of 1 - 0.000001(false positive)

0.99 of 0.000001 positive. So

$P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094$

Probability of a positive test and having the disease:

0.99 of 0.000001. So

$P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}$

What is the probability of actually having the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002$

0.00002 = 0.002% probability of actually having the disease

6 0

3 years ago