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Papessa [141]
2 years ago
10

Find the missing side length

Mathematics
1 answer:
Goshia [24]2 years ago
8 0

Answer: 13.86

Step-by-step explanation:

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The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand
Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                         P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean strength of 50 items = 74.28

            \sigma = population standard deviation = 2.15

            n = sample of items = 50

            \mu = population mean tensile strength after machine was adjusted

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                  significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                 = [ 74.28-1.96 \times {\frac{2.15}{\sqrt{50} } } , 74.28+1.96 \times {\frac{2.15}{\sqrt{50} } } ]

                 = [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>

8 0
3 years ago
Stretch your thinking rewrite this sentence using the word fewer: Carey reads 10 more pages than Lucey .
Free_Kalibri [48]
Lucy read 10 fewer pages than Carey
5 0
3 years ago
A box contains 8 yellow, 3 orange and 5 purple candies. A tray contains 7 red, 4 blue and 9 orange drinks. If Asia chooses one c
gayaneshka [121]
Okay so we have to first find the total.
8+3+5= 16 candies
7+4+9= 20 drinks
Since she wants to get a yellow candy and any drink but orange, put that information as a fraction.
8/16= 50% or 1/2
11/20= 55% (By the way you get 11 because of the 7 red and 4 blue).
And that's your answer!
Hope this helped and good luck! ^=^
7 0
3 years ago
In the figure below, what are the lengths of “a”, “b”, and “c”? Round answers to nearest tenth.
antoniya [11.8K]

Answer:

use a ruler to answer the question

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3 years ago
The time to complete a standardized exam is approximately normal with a mean of 70 minutes &amp; a standard deviation of 10 minu
Gennadij [26K]

if the mean is 70 minutes, then 90 minutes is 2 standard deviations above 70. if you look at the bell curve i drew, the percent of exams that took over 90 minutes is 2.5% <span>
</span>
7 0
3 years ago
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