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2 years ago

Find the parametric equation of the line r(t), 0 ≤ t ≤ 1, between P = (2, − 3, 1) and Q = (− 2, 1, 7).

Mathematics

1 answer:

alisha [4.7K]2 years ago

7 0

Answer:

x=-2+3t , y=-9-10t

Step-by-step explanation:

I'm going to start by solving the first equation for t....which will give (x-x1)/(x2-x1) =t.

You can now substitute this in the second equation to get....y=y1 + (y2-y1)(x-x1)/(x2-x1)

Rearranging, you get y-y1 = (y2-y1) / (x2-x1) * (x-x1) which is equivalent to y-y1= m(x-x1) which is the point-slope formula for a line.

So since t was varying between 0 and 1,

plugging in t=0 you get x=x1 and y = y1 givin you the coordinate (x1,y1)

similarly, plugging in t=1 leads to (x2,y2)

So this is a representation of a segment connecting these 2 points.

B) just plug in the points into your given formula from part a to get...

x=-2 +(1--2)t and y = 9 +(-1-9)t

or x=-2+3t , y=-9-10t

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Solve the following inequality. 21<_-3(x - 4) < 30 A. 28<_x < 37 B. -3 <_ x < 6 C. -6 < x <_ -3 D. -10 &

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Step-by-step explanation:

we have

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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of n is required, and n is found for M = 0.009. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$