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beks73 [17]
1 year ago
7

Is the expression below a perfect square trinomial or the difference of two squares?

Mathematics
1 answer:
DanielleElmas [232]1 year ago
7 0

The expression shown below is a difference of two squares.

<h3>Is a given expression a perfect square trinomial or a difference of two squares?</h3>

In this problem we have an algebraic expression that has to be checked by algebraic procedures. The complete procedure is shown below:

(x² + 8 · x + 16) · (x² - 8 · x + 16)                  Given

(x + 4)² · (x - 4)²                                           Perfect square trinomial

[(x + 4) · (x - 4)] · [(x + 4) · (x - 4)]                 Definition of power / Associative and commutative property

(x² - 16)²                                                      Difference of squares / Definition of power / Result

The expression shown below is a difference of two squares.

To learn more on differences of squares: brainly.com/question/11801811

#SPJ1

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Answer:

The  90%  confidence interval for population mean is   14.7  <  \mu  <  19.1

Step-by-step explanation:

From the question we are told that

   The sample mean is  \= x  =  16.9

    The confidence level is  C  =  0.90

     The sample size is  n  =  45

     The standard deviation

Now given that the confidence level is  0.90 the  level of significance is mathematically evaluated as

       \alpha =  1-0.90

       \alpha  =  0.10

Next we obtain the critical value of  \frac{\alpha }{2}  from the standardized normal distribution table. The values is  Z_{\frac{\alpha }{2} } =  1.645

The  reason we are obtaining critical values for \frac{\alpha }{2}  instead of  that of  \alpha  is because \alpha  represents the area under the normal curve where the confidence level 1 - \alpha (90%)  did not cover which include both the left and right tail while \frac{\alpha }{2}  is just considering the area of one tail which is what we required calculate the margin of error

  Generally the margin of error is mathematically evaluated as

        MOE  =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

         MOE  = 1.645*  \frac{ 9 }{\sqrt{45} }

         MOE  = 2.207

The  90%  confidence level interval is mathematically represented as

      \= x  -  MOE  <  \mu  <  \= x  +  MOE

substituting values

     16.9 -  2.207  <  \mu  <  16.9 +  2.207

    16.9 -  2.207  <  \mu  <  16.9 +  2.207

     14.7  <  \mu  <  19.1

         

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