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storchak [24]
3 years ago
6

Does anyone understand how to solve this?

Mathematics
1 answer:
yanalaym [24]3 years ago
3 0

You solve it by filling in the function value and simplifying the result.

a.

\displaystyle \frac{f(x)-f(a)}{x-a}=\frac{\left(3x^{2}+x+3\right)-\left(3a^{2}+a+3\right)}{x-a}\\\\=\frac{3\left(x^{2}-a^{2}\right)+\left(x-a\right)}{x-a}\\\\=\frac{3(x+a)(x-a)+(x-a)}{x-a}=3(x+a)+1

b.

\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{\left(3(x+h)^{2}+(x+h)+3\right)-\left(3x^{2}+x+3\right)}{h}\\\\=\frac{\left(3\left(x^{2}+2hx+h^{2}\right)+(x+h)+3\right)-\left(3x^{2}+x+3\right)}{h}\\\\=\frac{3\left(2hx+h^{2}\right)+h}{h}=6x+1+3h

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The data is missing in the question. The data is provided below :

Document : 1     2      3      4      5     6      7      8

Brand A       17  29    18    14    21   25    22    29

Brand B       21  38    15    19    22   30    31   37

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Null hypothesis : $h_A = h_B$

Alternate hypothesis : $h_A > h_B$

These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.

Significance level = 0.05

Therefore the standard error is :  $SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$

                                                         = 3.602

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$t=\frac{(x_1-x_2)-d}{SE}$

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        $s_2$ = standard deviation of the sample 2

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        $n_2$ = size of the sample 2

         $x_1$ = mean of the sample 1

        $x_2$ = mean of the sample 2      

          d = the hypothesis difference between the population mean

The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P($t \leq -1.32$) = 0.1042  

Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.

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