(6x)^2 can be thought of as a vertical stretch by a factor of<br> O 6<br> O 36<br> O nothing

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the

Darya [45]

Answer:

Brand 1 Brand 2 Difference

37734 35202 2532

45299 41635 3664

36240 35500 740

32100 31950 150

37210 38015 −805

48360 47800 560

38200 37810 390

33500 33215 285

Sum of difference = 2532+ 3664+740+150 −805+ 560 +390 +285 = 7516

Mean = $d=\frac{7516}{8}$

Mean = $d=939.5$

a) d= 939.5

$\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}$

$=\sqrt{\dfrac{(2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2}{8-1}}$

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So, $t_{\frac{\alpha}{2}}=3.499$

Formula for confidence interval $= \left( \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}} \right)$

Substitute the values

confidence interval $= 939.5 \pm 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

confidence interval $= 939.5 - 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$ to $= 939.5 + 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

Confidence interval $−843.396\$ to $2722.396$

8 0

3 years ago

Can somebody help me with 5-10

NeX [460]

Ok so use your fingers you have ten and take five away from your fingers or you can us a calculator but the answer is 5

3 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$