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3 years ago

A segment can have more than one bisector. True False

1 answer:

4 0

<span>Can a Segment have more than one bisector. Yes A segment can have more than one bisector. For every line segment, there is one perpendicular bisector that passes through the midpoint. There are infinitely many bisectors, but only one perpendicular bisector for any segment.</span>

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Two sisters, sister A and sister B, play SCRABBLE with each other every evening. Sister A is a statistician, and she draws a ran

snow_tiger [21]

Answer:

First question: LCL = 522, UCL = 1000.5

Second question: A sample size no smaller than 418 is needed.

Step-by-step explanation:

First question:

Lower bound:

0.36 of 1450. So

0.36*1450 = 522

Upper bound:

0.69 of 1450. So

0.69*1450 = 1000.5

LCL = 522, UCL = 1000.5

Second question:

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 $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

The project manager believes that p will turn out to be approximately 0.11.

This means that $\pi = 0.11$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The project manager wants to estimate the proportion to within 0.03

This means that the sample size needed is given by n, and n is found when M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.11*0.89}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.11*0.89}$

$\sqrt{n} = \frac{1.96\sqrt{0.11*0.89}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.11*0.89}}{0.03})^2$

$n = 417.9$

Rounding up

A sample size no smaller than 418 is needed.

6 0

3 years ago

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$