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1 month ago

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There are 130 people in a sport centre. 73 people use the gym. 62 people use the swimming pool. 58 people use the track. 22 peop

le use the gym and the pool. 29 people use the pool and the track. 25 people use the gym and the track. 11 people use all three facilities. A person is selected at random. What is the probability that this person doesn't use any facility

1 answer:

zzz [12.3K]1 month ago

6 0

p(the individual does not use any facilities)= 59/130

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Answer:

The maximum distance from the base camp at the conclusion of the third displacement is 6.69 km

Step-by-step explanation:

Each displacement can be represented as a vector, defined by magnitude and direction.

Vectors can be expressed in terms of their x and y coordinates as shown

$\vec{t}=(x, y)$

For displacements a and c, their vector coordinates would be:

$\vec{a}=(2, 0)$

$\vec{c}=(-1, 0)$

Given that displacement b is oriented at 30° north from due east, we can determine its x and y coordinates by applying the following formulas:

$x=(magnitude)*cos(angle)$

$y=(magnitude)*sin(angle)$

Note: the angle applied in the formula is that which is formed with the east point measured counterclockwise.

Thus, the x and y coordinates for displacement b would be:

$\vec{b}=(2*cos(30), 2*sin(30))$

As vector addition is commutative, the arrangement of displacements will not influence the final position; however, any directional alteration in a displacement will affect the resultant position. To ascertain the greatest distance, we ought to compute several combinations and identify the one yielding the largest magnitude:

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

Each resultant vector can be determined by summing up the respective components. Next, use the following formula to find the magnitude:

$|\vec{R}|=\sqrt[ ]{(R_{x})^2 +{(R_{y})^2}}$

Now, let’s execute the calculations!

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$R_{1_x}} =2+2*cos(30)-1=2.73$

$R_{1_y}} =0+2*sin(30)+0=1$

$\vec{R_{1}}=(2.73,1)$

$|\vec{R_{1}}|=\sqrt[ ]{(2.73)^2 +{(1)^2}}=3.86$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$R_{2_x}} =2-2*cos(30)-1=0.73$

$R_{2_y}} =0-2*sin(30)+0=-1$

$\vec{R_{2}}=(-0.73,-1)$ }

$|\vec{R_{2}}|=\sqrt[ ]{(-0.73)^2 +{(-1)^2}}=1.03$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$R_{3_x}} =2+2*cos(30)+1=4.73$

$R_{3_y}} =0+2*sin(30)-0=1$

$\vec{R_{3}}=(4.73,1)$

$|\vec{R_{3}}|=\sqrt[ ]{(4.73)^2 +{(1)^2}}=6.69$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

$R_{4_x}} =2-2*cos(30)+1=1.26$

$R_{4_y}} =0-2*sin(30)-0=1$

$\vec{R_{4}}=(1.26,-1)$

$|\vec{R_{4}}|=\sqrt[ ]{(1.26)^2 +{(-1)^2}}=1.79$

So, after performing all calculations, we can confirm that the vector $\vec{R_{3}}$ has the maximum magnitude. Therefore, the maximum distance possible from the base camp after the third displacement is 6.69 km

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