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Mkey [24]
3 years ago
7

What is the product of (x 1)^2?

Mathematics
1 answer:
Viktor [21]3 years ago
6 0
The product of (x1)^2 is 20
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Whích model represents a function?<br> A)<br> B)<br> C)
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Where is the model on the question ?
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Simplify completely quantity x squared minus 3 x minus 54 over quantity x squared minus 18 x plus 81 times quantity x squared pl
oee [108]
Your statemtent is incomplete.

I found the samestatment with the complete words: <span>Simplify completely quantity x squared minus 3 x minus 54 over quantity x squared minus 18 x plus 81 times quantity x squared plus 12 x plus </span>36 over x plus 6

Given that your goal is to learn an be able to solve any similar problem, I can teach you assuming that what I found is exactly what you need.

x^2 - 3x - 54            x^2 + 12x + 36
------------------    x   ---------------------
x^2 - 18x + 81              x + 6

factor x^2 - 3x - 54 => (x - 9)(x + 6)

factor x^2 - 18x + 81 => (x - 9)^2

factor x^2 + 12x + 36 = (x + 6)^2

Now replace the polynomials with the factors=>

(x - 9) (x + 6) (x + 6)^2        (x + 6)^2       x^2 + 12x + 36
------------------------------ =  --------------- = --------------------
    (x - 9)^2 (x + 6)                 (x - 9)               x - 9

4 0
3 years ago
Read 2 more answers
Pls help me on this question.<br>Do NOT answer 13 and 15. Only answer 14.
Ilia_Sergeevich [38]
1,700 good luck hope it helps you 100*17=1,700
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3 years ago
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A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti
jeyben [28]

Answer:

We need a sample of size at least 13.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

\pi = \frac{0.438 + 0.642}{2} = 0.54

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}

0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}

\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}

n = 12.21

Rounding up

We need a sample of size at least 13.

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3 years ago
Someone please help me, I need the type of statement and reason.
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The type of statement is a congruent
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