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Lynna [10]
3 years ago
15

HELP PLEASEE !!! What is the left-hand limit of f(x)=|x-3|/x-3 as approaches 3?

Mathematics
2 answers:
aleksklad [387]3 years ago
6 0

Answer:

-1

Step-by-step explanation:

x-3 is negative when x aproaches 3 from the left so the limit can be rewritten as

\lim_{x \to 3-} \frac{-x+3}{x-3}\\ = \lim_{x \to 3-} \frac{-(x-3)}{x-3}\\\\=-1

ladessa [460]3 years ago
3 0

Answer:

Option D  -1

Step-by-step explanation:

Given : Function f(x)=\frac{|x-3|}{x-3}

To find : What is the left-hand limit of function as approaches 3?      

Solution :

Function f(x)=\frac{|x-3|}{x-3}

In left-hand limit, x\rightarrow 3^-

x

So, f(x)=\frac{|x-3|}{x-3}=\frac{-(x-3)}{x-3}=-1,x\neq3

\lim_{x\rightarrow3^-}f(x)= \lim_{x\rightarrow3^-} (-1)=-1

Therefore, The left-hand limit of function as approaches 3 is -1.

So, Option D is correct.

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<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size  'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample  S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size  n₂ = 24</em>

<em>mean of the second sample  "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample  S₂ =  2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em>    </em>t = \frac{x^{-} _{1} -  x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} }  }

<em>    where         </em>S^{2}   = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2}   }{n_{1} +n_{2} -2}

                      S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}

<em>              substitute values and we get  S² =  40.988</em>

<em>     </em>t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24}  )}<em></em>

<em>  </em>   t =  1.3622

  Calculated value t = 1.3622

Tabulated value 't' =  2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em>      </em>

<em>                        </em>

<em>                    </em>

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