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yawa3891 [41]
3 years ago
10

Which polynomials are prime? Check all that apply.

Mathematics
2 answers:
FromTheMoon [43]3 years ago
10 0
The awnser is A,D,and E
KIM [24]3 years ago
6 0

Answer:

The prime polynomials are 1, 4 and 5

Step-by-step explanation:

Given some polynomials we have to classify the polynomials prime or not.

Prime polynomials are the polynomial with integer coefficients that cannot be factored into lower degree polynomials.

15x^2 + 10x - 9x + 7

⇒ 5(3x+2)-(9x-7)

can't be factored into lower degree polynomial ∴ prime polynomial.

20x^2 - 12x + 30x - 18

⇒ 4x(5x-3)+6(5x-3)

⇒ (4x+6)(5x-3)

hence, not a prime polynomial.

6x^3 + 14x^2 - 12x - 28

⇒ 2x^2(3x+7)-4(3x+7)

⇒ (2x^2-4)(3x+7)

hence, not a prime polynomial.

8x^3 + 20x^2 + 3x + 12

⇒ 4x^2(2x+5)+(3x+20)

can't be factored into lower degree polynomial ∴ prime polynomial.

11x^4 + 4x^2 - 6x^2 - 16

⇒ x^2(11x^2+4)-2(3x^2+8)

can't be factored into lower degree polynomial ∴ prime polynomial.

The prime polynomials are 1, 4 and 5

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katrin [286]

Answer:

-½(30) = -15

Step-by-step explanation:

Marta's pool decrees ½ inch each day, so that would be -½ each day. You would take that -½ and multiply it by 30 so you can find out the decrees in inches over 30 days so the equation would be -½ × 30 = -15 (im sorry if that didn't make sense)

4 0
2 years ago
An industrial psychologist conducted an experiment in which 40 employees that were identified as "chronically tardy" by their ma
iren [92.7K]

Answer:

The probability plot of this distribution shows that it is approximately normally distributed..

Check explanation for the reasons.

Step-by-step explanation:

The complete question is attached to this solution provided.

From the cumulative probability plot for this question, we can see that the plot is almost linear with no points outside the band (the fat pencil test).

The cumulative probability plot for a normal distribution isn't normally linear. It's usually fairly S shaped. But, when the probability plot satisfies the fat pencil test, we can conclude that the distribution is approximately linear. This is the first proof that this distribution is approximately normal.

Also, the p-value for the plot was obtained to be 0.541.

For this question, we are trying to check the notmality of the distribution, hence, the null hypothesis would be that the distribution is normal and the alternative hypothesis would be that the distribution isn't normal.

The interpretation of p-valies is that

When the p-value is greater than the significance level, we fail to reject the null hypothesis (normal hypothesis) and but if the p-value is less than the significance level, we reject the null hypothesis (normal hypothesis).

For this distribution,

p-value = 0.541

Significance level = 0.05 (Evident from the plot)

Hence,

p-value > significance level

So, we fail to reject the null or normality hypothesis. Hence, we can conclude that this distribution is approximately normal.

Hope this Helps!!!

7 0
3 years ago
Define a variable and write each phrase as an algebraic expression
asambeis [7]
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Area = (side)^2

Since the square of the square's side is its total area, then if vice versa, we have to find the square root of the area to find the length. Thus,

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3 years ago
Select the correct answer.
Setler [38]
A or D is the answer
4 0
1 year ago
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