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Question Number 169774 by mr W last updated on 08/May/22

Commented by cortano1 last updated on 08/May/22

$w h a t t h i s a n s w e r ?$
Commented by mr W last updated on 08/May/22

Commented by mr W last updated on 08/May/22

$p o i n t C w h i c h f u l f i l l s A C + B C ⩽ 10$ $m u s t l i e o n o r i n s i d e t h e e l l i p s e$ $w i t h A a n d B a s f o c i . s o t h e p r o b a b i l i t y$ $i s p = \frac{a r e a o f g r e e n e l l i p s e}{a r e a o f r e d s q u a r e} .$ $e q n . o f e l l i p s e i s \frac{x^{2}}{5^{2}} + \frac{y^{2}}{3^{2}} = 1, i t s a r e a$ $i s π \times 3 \times 5 = 15 π .$ $t h e a r e a o f s q u a r e i s 20^{2} = 400 .$ $\Rightarrow p = \frac{15 π}{400} = \frac{3 π}{80} \approx 11 . 8 %$
	Answered by cortano1 last updated on 08/May/22
	${ ((AC=(√((x+4)^2 +y^2 )))),((BC=(√((x−4)^2 +y^2 )))) :} ⇒AC+BC =(√((x+4)^2 +y^2 )) +(√((x−4)^2 +y^2 )) ≤10 (x+4)^2 +y^2 ≤100+(x−4)^2 +y^2 −20(√((x−4)^2 +y^2 )) 16x ≤100 −20(√((x−4)^2 +y^2 )) 5(√((x−4)^2 +y^2 )) ≤ 25−4x 25x^2 −200x+400+25y^2 ≤625−200x+16x^2 9x^2 +25y^2 ≤225 (x^2 /(25)) + (y^2 /9) ≤1 ⇒p(A)= ((π(5)(3))/(400)) = ((3π)/(80))$
	${\begin{cases} A C = \sqrt{{(x + 4)}^{2} + y^{2}} \\ B C = \sqrt{{(x - 4)}^{2} + y^{2}} \end{cases}$ $\Rightarrow A C + B C = \sqrt{{(x + 4)}^{2} + y^{2}} + \sqrt{{(x - 4)}^{2} + y^{2}} ⩽ 10$ ${(x + 4)}^{2} + y^{2} ⩽ 100 + {(x - 4)}^{2} + y^{2} - 20 \sqrt{{(x - 4)}^{2} + y^{2}}$ $16 x ⩽ 100 - 20 \sqrt{{(x - 4)}^{2} + y^{2}}$ $5 \sqrt{{(x - 4)}^{2} + y^{2}} ⩽ 25 - 4 x$ $25 x^{2} - 200 x + 400 + 25 y^{2} ⩽ 625 - 200 x + 16 x^{2}$ $9 x^{2} + 25 y^{2} ⩽ 225$ $\frac{x^{2}}{25} + \frac{y^{2}}{9} ⩽ 1 \Rightarrow p (A) = \frac{π (5) (3)}{400} = \frac{3 π}{80}$
	Commented by mr W last updated on 08/May/22

	$v e r y n i c e!$
	Commented by cortano1 last updated on 08/May/22

	$m y a n s w e r c o r r e c t ?$
	Commented by mr W last updated on 08/May/22

	$y e s$
	Commented by cortano1 last updated on 08/May/22

	$t h a n k s s i r$
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