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balu736 [363]
3 years ago
11

The equation 24x2+25x−47ax−2=−8x−3−53ax−2 is true for all values of x≠2a, where a is a constant.

Mathematics
1 answer:
Yuki888 [10]3 years ago
6 0

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have:

24x2+25x−47=(−8x−3)(ax−2)−53

You should then multiply (−8x−3) and (ax−2) using FOIL.

24x2+25x−47=−8ax2−3ax+16x+6−53

Then, reduce on the right side of the equation

24x2+25x−47=−8ax2−3ax+16x−47

Since the coefficients of the x2-term have to be equal on both sides of the equation, −8a=24, or a=−3.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal.

The answer is B

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3 0
3 years ago
Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th
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Answer:

a

\= x  =  40.85

b

E = 5.85

Ca  

   t_c =  2.08

Cb

   t_c  =   1.282

Explanation:

From the question we are told that

     The sample size is n  =  100

     The upper limit of the 95% confidence interval is  b =  47.2 years

     The lower limit of the 95% confidence interval is   a =  34.5 years

Generally the sample mean is mathematically represented as

         \= x  =  \frac{a + b }{2}

=>    \= x = \frac{47.2 + 34.5 }{2}

=>    \= x  =  40.85

Generally the margin of error is mathematically represented as

         E =  \frac{b- a }{ 2}

=>      E =  \frac{47.2- 34.5 }{ 2}

=>      E = 5.85

Considering question C a  

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

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

The  sample size is  n =  22

Given that the sample size is not sufficient enough i.e n < 30 we will make use of the student t distribution table  

Generally the degree of freedom is mathematically represented as

           df =  n- 1

=>        df =  22 - 1

=>        df =  21

Generally from the student  t  distribution table the critical value  of  \frac{\alpha }{2} at a degree of freedom  of  21 is  

   t_c =t_{\frac{\alpha }{2} ,  21  } =  2.08

Considering question C b

From the question we are told the confidence level is  80% , hence the level of significance is    

      \alpha = (100 - 80 ) \%

=>   \alpha = 0.20

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

   t_c  =Z_{\frac{\alpha }{2} } =  1.282

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