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mylen [45]
4 years ago
14

Which two answers???

Mathematics
1 answer:
Norma-Jean [14]4 years ago
6 0

Answer:

B and D

Step-by-step explanation:

have a nice day and mark me brainliest! :)

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Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original pri
Thepotemich [5.8K]

Answer:

hgngvhynghvh

Step-by-step explanation:

3 0
3 years ago
According to the synthetic division below, which of the following statements are true?
makkiz [27]

Answer:

Correct options are A and D

Step-by-step explanation:

According to the synthetic division in the diagram you can write down the result of division:

2x^2+9x-7=(x-(-6))(2x-3)+11,\\ \\2x^2+9x-7=(x+6)(2x-3)+11.

Therefore,

  • when 2x^2+9x-7 is divided by x+6, the remainder is 11 (option D is correct). To find the remainder after division by x-6, you have to use another synthetic division. Actually, 2x^+9x-7=(x-6)(2x+21)+119, then the remainder is 119 (option C is false).
  • when x=-6, the expression x+6 is -6+6=0 and 2x^2+9x-7=0\cdot (2x-3)+11=11 (option A is correct). You cannot state the same when x=6 (option B is false).
  • neither x-6 nor x+6 is a factor of 2x^2+9x-7, because the remainders in both cases are not equal to 0 (options E and F are false).

4 0
4 years ago
Read 2 more answers
In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume
kobusy [5.1K]

Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by \phi . The values of \phi can be found in the attached file.

P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.  

Download pdf
6 0
4 years ago
I need help with math!!! time is running out!!! will mark brainliest!!!
sashaice [31]
1.

Answer B

2.

Answer D.

3.

Answer D.

4.

Answer C.
8 0
3 years ago
What is the factored form of x2 – 7x + 12?
const2013 [10]

Answer:

D

Step-by-step explanation:

Given

x² - 7x + 12

Consider the factors of the constant term (+ 12) which sum to give the coefficient of the x- term (- 7)

The factors are - 4 and - 3, since

- 4 × - 3 = + 12 and - 4 - 3 = - 7, thus

x² - 7x + 12 = (x - 4)(x - 3) ← in factored form → D

4 0
3 years ago
Read 2 more answers
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