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jeyben [28]
3 years ago
13

Click on the photo! Needing help ASAP please!♥️

Mathematics
2 answers:
Rufina [12.5K]3 years ago
8 0
<h3>Answer: Choice A)  x^4+20x^2-100</h3>

=============================================================

Explanation:

Choice B has the GCF x we can factor out like so

10x^4-5x^3+70x^2+3x = x(10x^3-5x^2+70x+3)

Showing that choice B is <u>not</u> prime. If a polynomial can be factored, then we consider it not prime. It's analogous to saying a number like 15 isn't prime because 15 = 3*5, ie 15 can be factored into something that doesn't involve 1 as a factor.

In contrast, we consider 7 prime because even though 7 = 1*7, there aren't any other ways to write this integer as a factorization if we don't involve 1.

-----------------------------------

Choice C is a similar story. This time we can factor out 3

3x^2 + 18y = 3(x^2 + 6y)

So we can rule this out as well.

-----------------------------------

Choice D is a bit tricky, but we can use the difference of cubes factoring rule

a^3 - b^3 = (a-b)(a^2+ab+b^2)

where in this case a = x and b = 3y^2

Note how b^3 = (3y^2)^3 = 3^3*(y^2)^3 = 27y^(2*3) = 27y^6

All of this means choice D can be factored and it's not prime either.

------------------------------------

We've ruled out choices B through D. The answer must be choice A.

If you let w = x^2, then w^2 = x^4

The polynomial w^2+20w-100 is prime because setting it equal to zero and solving for w leads to irrational solutions. I'm assuming your teacher wants you to factor over the rational numbers.

Because w^2+20w-100 can't be factored over the rational numbers, neither can x^4+20x^2-100. This confirms that choice A is prime.

Lana71 [14]3 years ago
5 0
A
This is like a base form of a prime polynomial
X^2+x+1
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3 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
2 years ago
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