(Geometry) The answers are x = 15 and y = 9 but I don't know how. Help me understand this

Mathematics

1 answer:

Kryger [21]3 years ago

4 0

Check the picture below.

$\bf \stackrel{\measuredangle DAC}{4y+2x}~~=~~\stackrel{\measuredangle BCA}{9y-x}\implies 2x=5y-x\implies 3x=5y\implies \boxed{x=\cfrac{5y}{3}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\measuredangle DAB}{[(4y+2x)+35]}~~+~~\stackrel{\measuredangle ADC}{5x+4}~~=~~180\implies 4y+2x+5x+39 = 180 \\\\\\ 4y+7x+39=180\implies 4y+7x=141\implies \stackrel{\textit{substituting "x"}}{4y+7\left( \boxed{\cfrac{5y}{3}} \right)} = 141$

$\bf 4y+\cfrac{35y}{3}=141\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( 4y+\cfrac{35y}{3} \right)=3(141)}\implies 12y+35y=423 \\\\\\ 47y=423\implies y=\cfrac{423}{47}\implies \blacktriangleright y = 9 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{5y}{3}}\implies x = \cfrac{5(9)}{3}\implies \blacktriangleright x = 15 \blacktriangleleft$

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<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

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<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$