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mezya [45]
3 years ago
5

How quickly do synthetic fabrics such as polyester decay in landfills? A researcher buried polyester strips in thes oil for diff

erent lengths of time, then dug up the strips and measured the force required to break them. Breaking strength is easy to measure and is a good indicator of decay. Lower strength means the fabric is decayed. In part of the study researchers buried 10 strips of polyester fabric in well-drained soil in the summer. Five of the strips, chosen at random, were dug up after 2 weeks; the other 5 were dug up after 16 weeks. . Here are the data (in pounds of breaking strength): Sample 1 (2 weeks) 118 126 126 120 129 Sample 2 (16 weeks) 124 98 110 140 110 These data can be downloaded: Minitab data Excel or text data Find a 95% confidence interval for the difference in the mean breaking strengths, specifically the mean 2-week breaking strength minus the mean 16-week breaking strength.
Mathematics
1 answer:
ad-work [718]3 years ago
5 0

Answer:

no enough evidence to draw up a conclusion

Step-by-step explanation:

Data:

let: \mu _{1} = decay after 2 weeks

    \mu _{2} = decay after 16 weeks

Making the hypotheses:

Null hypothesis; H_{o} = \mu _{1}  - \mu _{2} = 0

the deviation = 0.05

the standard difference  = 4

Conditions:

The normal distribution curve can be plotted on a graph and the plot shows that the distribution is a skewed distribution.

Then,

t = \frac{(123.8-16.4)-0}{\sqrt{\frac{4.6^{2} }{5}+\frac{16.09^{2} }{5}   } } \\  = 0.989

Therefore, it can be concluded that 0.1983 > 0.05. This presents our failure to reject the null hypothesis.

Thus, there is not enough evidence to conclude that polyester decays more in less than 2 weeks.

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Find the area between the graph of the function and the x-axis over the given interval, if possible.
vodomira [7]

Answer:

A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

A = \int_{-\infty}^1 \frac{5}{(x-2)^2}

For this case we can use the following substitution u = x-2 and we have dx = du

A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du

And if we solve the integral we got:

A= -\frac{5}{u} \Big|_a^b

And we can rewrite the expression again in terms of x and we got:

A = -\frac{5}{x-2} \Big|_{-\infty}^1

And we can solve this using the fundamental theorem of calculus like this:

A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5

So then the integral converges and the area below the curve and the x axis would be 5.

7 0
3 years ago
The equation of the line that contains diagonal HM is y = 2/3 x + 7.
nirvana33 [79]

Answer:

The slope of the line that contains diagonal OE will be = -3/2

Step-by-step explanation:

We know the slope-intercept form of the line equation

y = mx+b

Where m is the slope and b is the y-intercept

Given the equation of the line that contains diagonal HM is y = 2/3 x + 7

y = 2/3 x + 7

comparing the equation with the slope-intercept form of the line equation

y = mx+b

Thus, slope = m = 2/3

  • We know that the diagonals are perpendicular bisectors of each other.

As we have to determine the slope of the line that contains diagonal OE.

As the slope of the line that contains diagonal HM = 2/3

We also know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line.

Therefore, the slope of the line that contains diagonal

OE will be = -1/m = -1/(2/3) = -3/2

Hence, the slope of the line that contains diagonal OE will be = -3/2

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Paraphin [41]

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