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Ganezh [65]
3 years ago
12

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

ard 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. Find a 95% confidence interval for the mean of tensile strength after the machine was adjusted. Does this data suggest that the tensile strength was changed after the adjustment
Mathematics
1 answer:
Elanso [62]3 years ago
8 0

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>

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We know that at 5 am, it was -7 degrees. Since the change of temperature is a negative number, then we subtract the absolute value of the change from -7. -7 - 3 = -10. Following this pattern, we then have:
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3 years ago
Read 2 more answers
Help asap! 50 Points! 10 Questions!
tensa zangetsu [6.8K]
Part 1) <span>If a cylinder has a surface area of 36 square inches and its radius is 1.6 inches, how tall is the cylinder in inches? π = 3.14. Round to the nearest tenth.

we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height

area of the base==pi*r</span>²*h-----> 3.14*1.6²*h----> 8.0384*h in²

perimeter of the base=2*pi*r---> 2*3.14*1.6----> 10.048 in

surface area=36 in²
so
36=2*[8.0384*h]+10.048*h----> 36=26.1248*h
h=36/26.1248----> h=1.38 in------> h=1.4 in

the answer Part 1) is
h=1.4 in

Part 2) <span>2. A regular hexagonal pyramid has a base area of 29 in^2 and a lateral area of 210 in^2. What is the surface area of the hexagonal pyramid?

surface area of the hexagonal pyramid=2*area of base+lateral area
</span>surface area of the hexagonal pyramid=2*29+210----> 268 in²

the answer part 2) is 
surface area=268 in²

Part 3)<span> What is the surface area of the following cylinder in square inches? Use π = 3.14 and round your answer to the nearest tenth.
Radius = 3.2 Height = 4.1
</span>we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height

area of the base==pi*r²*h-----> 3.14*3.2²*4.1----> 131.83 in²

perimeter of the base=2*pi*r---> 2*3.14*3.2----> 20.096in

surface area=2*131.83+20.096*4.1------> 346.05 in²-----> 346.1 in²

the answer Part 3) 
surface area=346.1 in²

part 4)<span>A cube has surface area of 150 in^2. How long is each side of the cube?

we know that
surface area of cube=6*b</span>²
where b is the length side of the cube
SA=6*b²------> b=√SA/6------> b=√150/6-----> √25----> b=5 in

the answer Part 4) is
b=5 in

Part 5)  <span>A cube has sides that are 3 in. long. What is the surface area of the cube?
</span>we know that
surface area of cube=6*b²
where b is the length side of the cube
b=3 in
so
SA=6*3²----> 54 in²

the answer Part 5) is
SA=54 in²

Part 6) <span>What is the surface area of this box?
Height = 3.5 Length = 6 width = 3

surface area of the box=2*area of the base+perimeter of base*height 

area of the base=6*3----> 18 ft</span>²

perimeter of the base=2*[6+3]----> 18 ft
height=3.5 ft

surface area of the box=2*18+18*3.5----> 99 ft²

the answer Part 6) is
SA=99 ft²

Part 7) <span>If a cylinder has a surface area of 59 square inches and its radius is 2.1 inches, how tall is the cylinder in inches? Use π = 3.14. Round to the nearest tenth
</span>

we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height

area of the base==pi*r²*h-----> 3.14*2.1²*h----> 13.8474*h in²

perimeter of the base=2*pi*r---> 2*3.14*2.1----> 13.188 in

surface area=59 in²
so
59=2*[13.8474*h]+13.188*h----> 59=40.8828*h
h=59/40.8828----> h=1.44 in------> h=1.4 in

the answer Part 7) is
h=1.4 in

Part 8) <span>What is the surface area of the cone below, rounded to the nearest tenth? Use π = 3.14.
Height = 12 Radius = 5
surface area of the cone=pi*r</span>²+pi*r*l
r=5 in
h=12 in
l=slant height
l²=r²+h²----> l²=5²+12²----> l²=169-----> l=13 in

surface area of the cone=3.14*5²+3.14*5*13-----> 282.6 in²

the answer Part 8) is
SA=282.6 in²

Part 9) <span>A square pyramid has a surface area of 202 ft^2 and a base area of 53 ft^2. What is the lateral area of the pyramid?

we know that
surface area=area of the base+lateral area
lateral area=surface area-area of the base
lateral area=202-53----> 149 ft</span>²

the answer Part 9) is
LA=149 ft²

Part 10)<span>What is the surface area of the cone below, rounded to the nearest tenth? Use π = 3.14
Radius = 2 Height = 4
</span>
surface area of the cone=pi*r²+pi*r*l
r=2 cm
h=4 cm
l=slant height
l²=r²+h²----> l²=2²+4²----> l²=20-----> l=4.47 cm

surface area of the cone=3.14*2²+3.14*2*4.47-----> 40.63 cm²

the answer Part 10) is
SA=40.6 cm²
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What is the value of the 24th term is the fourth term is 6 and the eighth term is 8
Alina [70]

The value of the 24th term is 16.

<h3>Arithmetic progression:</h3>
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where

a = first term

d = common difference

n = number of terms

Therefore,

6 = a + 3d

8 = a + 7d

let's find a and d

2 = 4d

d = 2 / 4

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a = 6 - 3 /2

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24th term = 4.5 + 23(0.5)

24th term = 4.5 + 11.5

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