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Rom4ik [11]
3 years ago
9

Show all work and provide necessary descriptions

Mathematics
1 answer:
masya89 [10]3 years ago
5 0

Let

P=(P_x,P_y),\quad Q=(Q_x,Q_y)

If M is the midpoint, the x and y coordinates of M are the average of the x and y coordinates of P and Q:

M=\left(\dfrac{P_x+Q_x}{2},\ \dfrac{P_y+Q_y}{2}\right)

We can solve this expression for the coordinates of Q:

M_x = \dfrac{P_x+Q_x}{2} \implies Q_x = 2M_x-P_x

M_y = \dfrac{P_y+Q_y}{2} \implies Q_y = 2M_y-P_y

Plug in the values for the coordinates of M and P to get

Q_x = 2M_x-P_x = 2\cdot 5-11 = 10-11=-1

Q_y = 2M_y-P_y = 2\cdot (-2) - (-10) = -4+10=6

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Pat will pass the exam if x ≥ 10. The probability that Pat will pass is:

P(pass)=P(x=10)+P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)+P(x=16)+P(x=17)+P(x=18)+P(x=19)+P(x=20)

The probability for each number of success is:

P(x=10)=\frac{20!}{(20-10)!10!}*0.25^{10}*0.75^{10}=0.0099\\\\P(x=11)= \frac{20!}{(20-11)!11!}*0.25^{11}*0.75^{9}=0.0030\\\\P(x=12)=\frac{20!}{(20-12)!12!}*0.25^{12}*0.75^{8}=0.00075\\\\P(x=13)=\frac{20!}{(20-13)!13!}*0.25^{13}*0.75^{7}=0.00015\\\\P(x=14)=\frac{20!}{(20-14)!14!}*0.25^{14}*0.75^{6}=0.0000257\\\\P(x=15)=\frac{20!}{(20-15)!15!}*0.25^{15}*0.75^{5}=3.426*10^{-6}\\\\

P(x=16)=\frac{20!}{(20-16)!16!}*0.25^{16}*0.75^{4}=3.569*10^{-7}\\\\P(x=17)=\frac{20!}{(20-17)!17!}*0.25^{17}*0.75^{3}=2.799*10^{-8}\\\\P(x=18)=\frac{20!}{(20-18)!18!}*0.25^{18}*0.75^{2}=1.555*10^{-9}\\\\P(x=19)=\frac{20!}{(20-19)!19!}*0.25^{19}*0.75^{1}=5.457*10^{-11}\\\\P(x=20)=\frac{20!}{(20-20)!20!}*0.25^{20}*0.75^{0}=9.095*10^{-13}\\\\

The probability that Pat will pass his exam is:

P(pass)=0.01386

4 0
3 years ago
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