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Bad White [126]
2 years ago
5

Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 6x + 9 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4

= 0 D. x2 + 5x − 4 = 0
Mathematics
2 answers:
aksik [14]2 years ago
6 0

A.\ x^2+6x+9=x^2+3x+3x+9=x(x+3)+3(x+3)=(x+3)(x+3)\\\\B.\ x^2-x-6=x^2+2x-3x-6=x(x+2)-3(x+2)=(x+2)(x-3)\\\\C.\ x^2+3x-4=x^2+4x-x-4=x(x+4)-1(x+4)=(x+4)(x-1)\\\\D.\ x^2+5x-4\leftarrow ANSWER

Hitman42 [59]2 years ago
4 0

The answer is D, because the others can be solved. A when factored is equal to -3, when B is factored it is equal to 3,-1, when C is factored it is 1,-4. D can be solved but you will only get irrational numbers which can't be used.

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garri49 [273]

Answer:

Step-by-step explanation:

In the normal distribution curve, the mean is in the middle and each line to the left and to the right of that mean represent 1- and 1+ the standard deviation.  If our mean is 400, then 400 + 50 = 450; 450 + 50 = 500; 500 + 50 = 550.  Going from the mean to the left, we subtract the standard deviation and 400 - 50 = 350; 350 - 50 = 300; 300 - 50 = 250.  We are interested in the range that falls between 350 and 450 as a percentage.  That range represents the two middle sections, each containing 34% of the data.  So the total percentage of response times is 68%.  We are looking then for 68% of the 144 emergency response times in town.  .68(144) = 97.92 or 98 emergencies that have response times of between 350 and 450 seconds.

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1 year ago
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Cameron wondered if the average score on a final exam was different between those who texted on a regular basis during the lectu
Margarita [4]

Answer:

c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

Step-by-step explanation:

Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class

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2 years ago
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Softa [21]
<span>m∠SYD = </span>106.02°
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pochemuha
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