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Free_Kalibri [48]
3 years ago
9

Helpppppppppppppppppppppppppppp

Mathematics
2 answers:
Masteriza [31]3 years ago
5 0
Nr B:
x = -7 +/- √(-64)
x = -7 +/- 8*√-1
x = -7 +/- 8i
inessss [21]3 years ago
5 0
In this case it is to find the roots of the polynomial.
 We have then:
 x ^ 2 + 14x + 17 = -96
 Rewriting:
 x ^ 2 + 14x + 113 = 0
 Applying resolver we have
 x = (- b +/- root (b ^ 2 - 4ac)) / (2a)
 Substituting values:
 x = (- (14) +/- root ((14) ^ 2 - 4 (1) (113))) / (2 (1))
 x = (- (14) +/- root ((196 - 452)) / (2 (1))
 x = (- 14 +/- root (-256))) / (2)
 x = (- 14 +/- i * 16)) / (2)
 x = (- 7 +/- i * 8)
 Answer:
 x = (- 7 +/- i * 8)
 (option 2)
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the rectangular gable is 8 feet tall the area of the gable is 120 square feet, what is the length of the base
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7 0
2 years ago
4 divided by 1/8 = <br> Please I’m taking math inventory lol
Angelina_Jolie [31]
Your answer is 32! 4 divided by 1/8 is 32.
6 0
3 years ago
-8 1/16 as decimal number
Anon25 [30]

Answer:

-8.0625

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which of the following is NOT a requirement of testing a claim about two population means when 1 and 2 are unknown and not assum
andrey2020 [161]

Answer:

The correct answer is D because they are unknown

Step-by-step explanation:

7 0
3 years ago
For a population with µ = 80 and σ = 20, the Sampling distribution of the Mean, based on n = 16 will have an expected value of t
allsm [11]

Answer:

Will have an expected value of the mean = 80 and a standard error of the mean = 5

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 80, \sigma = 20

n = 16

So the mean is 80 and the standard deviation is s = \frac{20}{\sqrt{16}} = 5

Will have an expected value of the mean = 80 and a standard error of the mean = 5

6 0
3 years ago
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