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1 year ago

Find the distance and displacement for the following figures :

Mathematics

1 answer:

Vinil7 [7]1 year ago

4 0

The figure (i) have a distance of 15.996 meters and a displacement of 13.892 meters and (ii) a distance of 480 centimeters and a displacement of 339.411 centimeters

<h3>How to find the distance and displacement in each trajectory</h3>

The distance is the sum of the lengths that form a trajectory and the displacement is the distance between the <em>initial</em> and <em>final</em> point of a trajectory. Then, we have the following results for each case:

Case I

Distance

d = 5 m + 0.5π · (7 m)

d ≈ 15.996 m

Displacement

$D = \sqrt{(12\,m)^{2}+(7 m)^{2}}$

D ≈ 13.892 m

Case II

Distance

d = 8 · (60 cm)

d = 480 cm

Displacement

$D =\sqrt{(240\,cm)^{2}+(240\,cm)^{2}}$

D ≈ 339.411 cm

To learn more on displacement: brainly.com/question/11934397

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Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

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(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

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Step-by step explanation:

Given X represents the number on die.

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For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

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$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

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$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

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A teacher has 3 hours to grade all the papers submitted by the 35 students in her class. She gets through the first 5 papers in

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So, she has 3hrs to grade all papers, for 35 students.. alrite.

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now, she has still 2 hours and a half, or 150 minutes, to do the remaining 30 papers... she has to work at a rate of 30/150 then... which is 1/5 simplified.

now if we take 1/6 as the 100%, what is 1/5 in percentage then?

$\bf \begin{array}{ccllll}
rate&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{1}{6}&100\\\\
\frac{1}{5}&x
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\\\\\\
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