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3 years ago

The difference of two numbers is 34 and their sum is 338

Mathematics

1 answer:

Assoli18 [71]3 years ago

4 0

X-y=34
x+y=338

We can use elimination in this problem.

2x=372
x=186

186+y=338
y=338-186
y=152

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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

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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 $2722.396$

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3 years ago

Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= he

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Answer:

$p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }$

g(t) = 0

And

The differential equation $9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$ is linear and homogeneous

Step-by-step explanation:

Given that,

The differential equation is -

$9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$

$e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0$

By comparing with y′+p(t)y=g(t), we get

$p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }$

g(t) = 0

And

The differential equation $9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$ is linear and homogeneous.

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