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zloy xaker [14]
3 years ago
12

HELP!!

Mathematics
1 answer:
leonid [27]3 years ago
6 0

Answer:

The focus point is (2 , 0) ⇒ answer D

Step-by-step explanation:

* Lets revise the equation of the parabola in standard form

- The standard form is (x - h)² = 4p(y - k)

- The focus is (h, k + p)

- The directrix is y = k - p

- If the parabola is rotated so that its vertex is (h , k) and its axis of

 symmetry is parallel to the x-axis, it has an equation of

 (y - k)² = 4p(x - h)

- The focus is (h + p, k)

- The directrix is x = h - p

* Lets solve the problem

∵ The equation of the parabola is y = 1/8(x² - 4x - 12)

- Lets make x² - 4x completing square

∵ √x² = x  

∴ The 1st term in the bracket is x

∵ 4x ÷ 2 = 2x

∴ The product of the 1st term and the 2nd term is 2x

∵ The 1st term is x

∴ the second term = 2x ÷ x = 2

∴ The bracket is (x - 2)²

∵  (x - 2)² = (x² - 4x + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out  

  the bracket to keep the equation as it

∴ (x² - 4x + 4) - 4 = (x - 2)² - 4

- Lets put the equation after making the completing square

∴ y = 1/8 [(x - 2)² - 4 - 12]

∴ y = 1/8 [(x - 2)² - 16] ⇒ multiply both sides by 8

∴ 8y = (x - 2)² - 16 ⇒ add 16 to both sides

∴ 8y + 16 = (x - 2)² ⇒ take from the left side 8 as a common factor

∴ 8(y + 2) = (x - 2)²

∴ The standard form of the equation of the parabola is

   (x - 2)² = 8(y + 2)

∵ The standard form of the equation is (x - h)² = 4p(y - k)

∴ h = 2 , k = -2 , 4p = 8

∵ The focus is (h , k + p)

∵ h = 2

∵ 4p = 8 ⇒ divide both sides by 4

∴ p = 2

∴ The focus = (2 , -2 + 2) = (2 , 0)

* The focus point is (2 , 0)

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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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