The correct formula to determine the volume of a cylinder is

shutvik [7]

The value of m2 is 1/7(m1 + 54)

3 0

3 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

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

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

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

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

= [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].

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.

8 0

3 years ago

Solve this indices a²b + a³b / a²b²

Zolol [24]

Step-by-step explanation:

Notice: The literal factors are all the combinations of a and b where the sum of the exponents is 4: a4, a³b, a²b², ab³, b4 ... The solution to the problem of the binomial coefficients without actually ... The upper index n is the exponent of the expansion; the lower index k indicates which term ... 1a5 + a4b + a3b² + a²b3 + ab4 + b5.

3 0

2 years ago

What is 3×1 and 3/5 A) 2 and 2/5 B) 4 and 3/5 C) 4 and 4/5 D) 7 and 4/5

Bas_tet [7]

Option C is your answer.

1 3/5 is the same as 8/5

3(8/5) = 24/5 = 4 and 4/5

3 0

3 years ago

A cone has a 4 radius and a 8 height what is the volume

Marta_Voda [28]

Hey there! I'm happy to help!

To find the volume of a cone, you multiply the base by the height and then divide by three.

First, we need to find the base. A cone has a circle for the base. To find the area of a circle, we square the radius and multiply it by pi (we will use 3.14).

4²=16

16×3.14= 50.24

Now, we multiply our base by the height.

50.24×8=401.92

Finally, we divide by three.

401.92÷3≈133.973 (to nearest thousandth)

Therefore, the volume of the cone is about 133.973 units cubed. If you were to use exact pi, it would be more like 134 because of all those decimals after pi!

Now you can find the volume of cones!

I hope that this helps! Have a wonderful day!

8 0

3 years ago