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jasenka [17]
2 years ago
14

GEOMETRY - FIND X ON THIS TRIANGLE

Mathematics
1 answer:
m_a_m_a [10]2 years ago
8 0

Answer: \frac{7\sqrt{3}}{2}

Step-by-step explanation:

\sin 60^{\circ}=\frac{x}{7}\\\\\frac{\sqrt{3}}{2}=\frac{x}{7}\\\\x=\frac{7\sqrt{3}}{2}

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Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is  = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = 1-\frac{_{n}^{12}\textrm{P}}{12^2}

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= \frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})

= 0.25 \times 0.618056 + 0.25 \times 0.996132 + 0.5 \times 1

= 0.903547

2).P( k = 10|at least 2 share their birthday in same month)

=P(k=10 and at least 2 born in the same month)/P(at least 2 share their birthday in same month)

= $0.25 \times \frac{0.996132}{0.903547}$

= 0.0.275617

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