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masya89 [10]
4 years ago
6

For what value of x is the rational expression below undefined?

Mathematics
2 answers:
Blizzard [7]4 years ago
4 0

Answer: The expression is undefined for x=4 and x=5.

The expression is undefined for any x that makes the denominator 0. This leads to solving a quadratic equation:

\frac{x+3}{x^2-9x+20}\\x^2-9x+20\neq 0\\x_{1,2}\neq\frac{9\pm\sqrt{9^2-80}}{2}=\frac{9\pm1}{2}\\x_1\neq4\\x_2\neq5

loris [4]4 years ago
3 0

Answer:

Step-by-step explanation:

Alright, lets get started.

The rational expression is gien as :

\frac{x+3}{x^2-9x+20}

For being the function undefined, the denominator must be equal to zero.

x^2 -9x +20=0

factoring

(x-5)(x-4)=0

This will give two values of x, which are

x = 5 and x = 4

So,the answer is 4 and 5.   :   Answer

Hope it will help :)

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Statistics professors believe the average number of headaches per semester for all students is more than 18. From a random sampl
kompoz [17]

Complete Question

Statistics professors believe the average number of headaches per semester for all students is more than 18. From a random sample of 15 students, the professors find the mean number of headaches is 19 and the standard deviation is 1.7. Assume the population distribution of number of headaches is normal.the correct conclusion at \alpha =0.001 is?

Answer:

There is no sufficient evidence to support the professor believe

Step-by-step explanation:

From the question we are told that

     The population mean is  \mu =  18

     The sample size is  n  =  15

      The sample mean is  \=  x  =  19

      The standard deviation is  \sigma =  1.7

      The level of significance is  \alpha  =  0.001

The null hypothesis is  H_o:  \mu = 18

The  alternative hypothesis is  H_a :  \mu > 18

 The critical value of the level of significance from the normal distribution table is    

         Z_{\alpha } =  3.290527

The test hypothesis is mathematically represented as

           t =  \frac{\= x  -  \mu }{ \frac{\sigma}{ \sqrt{n} } }

substituting values  

         t =  \frac{ 19  - 18}{ \frac{1.7}{ \sqrt{15} } }

         t =  2.28

Looking at the value of  t and  Z_{\alpha } we can see that t <  Z_{\alpha } so we fail to reject the null hypothesis.

This mean that there is no sufficient evidence to support the professor believe

     

       

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