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PSYCHO15rus [73]
3 years ago
7

A&b are complimentary angles a measures 32 what is the measure of b

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

B=58

Step-by-step explanation:

Complementary angles are angles where the sum equals 90°. So all you have to do is 90-32 and you get 58.

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Find the area of circle Y. The image is represented by circle Y with a sector XYZ that has an area of 85.5 square meters. That s
pochemuha

Answer:

  263.08 m²

Step-by-step explanation:

The fraction of circle area that a sector represents is proportional to the central angle.

<h3>Proportion</h3>

  circle area/circle angle = sector area/sector angle

  circle area / 360° = 85.5 m² / 117°

Multiplying by 360°, we have ...

  circle area = (360/117)(85.5 m²) = 263.08 m²

The area of the circle is about 263.08 m².

6 0
2 years ago
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
Please help me out on this one !!
Serjik [45]

I think the answer is c

5 0
3 years ago
Plsss help will give a good amount of points, need done asap
Alex
4+2i because you can’t have a negative square root
7 0
3 years ago
Read 2 more answers
(ノ◕ヮ◕)ノ*:・゚✧PLEQASE HELP QUICK!!!! What is the exact volume of cylinder with a height of 40 inches and a radius of 16 inches ___
Delicious77 [7]
I can't resist helping someone with magic <span>(ノ◕ヮ◕)ノ*:・゚✧

The formula for the volume of a cylinder is pi x r^2 x h.
That's 10240 x pi or 3</span><span>2169.91

</span>(ノ◕ヮ◕)ノ*:・゚✧ woosh you now have smol pener
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3 years ago
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