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masha68 [24]
3 years ago
15

Two cones have their heights in the ratio 1 : 3 and the radius of their bases in the ratio 3 : 1, show that their volumes are in

the ratio 3 : 1​
Mathematics
1 answer:
Nuetrik [128]3 years ago
4 0
<h2>\large\bold{\underline{\underline{Question:-}}}</h2>

Two cones have their heights in the ratio 1 : 3 and the radius of their bases in the ratio 3 : 1 show that their volumes are in the ratio 3 : 1.

<h2>\large\bold{\underline{\underline{Solution:-}}}</h2>

⇒ Ratio of heights of two cones = 1 : 3

⇒ Ratio of radius of their bases = 3 : 1

⇒ We know that V = 1/3πr²h

⇒ Ratio of their volume = V1 : V2

\underline{ \underline{ \sf{Let's \:  find  \: the  \: ratio  \: of \:  their  \: volumes:}}}

{ \implies{ \sf  {V _{1}  :  V_{2}}}}

{ \implies{ \sf{ \frac{1}{3} \pi \:  { r_{1}}^{2}  h_{1} :  \frac{1}{3}  \pi \:  { r_{2} }^{2}  h_{2}}}}

{ \implies{ \sf{ {r_{1}}^{2}  h_{1} :  { r_{2}}^{2} h_{2}}}}

{ \implies{ \sf{ \frac{ { r_{1} }^{2} }{ { r_{2} }^{2} }  :  \frac{ h_{2} }{ h_{1} } }}} \\

{ \implies{ \sf{ \frac{ {3}^{2} }{ {1}^{2} }  :  \frac{3}{1} }}} \\

{ \implies{ \sf{ \frac{9}{1} :  \frac{3}{1}  }}} \\

{ \implies{ \sf{3 : 1}}} \\

{ \therefore{ \sf{ \green{Ratio  \: of  \: their \:  volumes = 3 : 1}}}}

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Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

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