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vfiekz [6]
3 years ago
7

Which system is equivalent to 5y^2+6y^2=50

Mathematics
1 answer:
stepladder [879]3 years ago
3 0
5y²+6y²=50⇒11y²=50⇒y²=\frac{454}{100}⇒y=\sqrt{ \frac{454}{100} }⇒y=\frac{ \sqrt{454} }{10}
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Answer:

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Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

  • A man is appointed first and a woman is appointed next.
  • A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed)=P(MW)+P(WM)

=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

  • A man is appointed first and a woman is appointed next.
  • A woman is appointed first and a man is appointed next.
  • Two women are appointed

P(at least one woman is appointed)=P(MW)+P(WM)+P(WW)

P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}

In Part B, P(MW)+P(WM)=\frac{8}{39}

Therefore:

P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}

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