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borishaifa [10]
2 years ago
9

Considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we pr

esent their model. For our purposes, it can be described as a multinomial model with the four categories C1, C2, C3 , and C4 . For a sample of size n, let X = (X1, X2, X3 , X4 )′ denote the observed frequencies of the four categories. Hence, n = +4 i=1 Xi. The probability model is
Mathematics
1 answer:
Brrunno [24]2 years ago
4 0

Answer:

im not sure

Step-by-step explanation:

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a musician's hair was originally 3 inches long she asked the hairdresser to cut 5/6 of it off how many inches did she have cut o
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It would be half an inch long because there is 6 halves in 3 so 5/6 of it would be half an inch
5 0
3 years ago
Allie and Ezra are sitting on flagpoles, throwing beanbags at each other. Allie's flagpole is $35$ feet tall. Ezra's flagpole is
Vedmedyk [2.9K]

Ezra has to throw a beanbag 17 feet far to reach Allie

According to the question,

The distance between Allie and Ezra's flagpoles is 15 feet

length of Allie's flagpole = 35 feet

length of Ezra's flagpole = 27 feet

Difference between Allie and Ezra's flagpole = (35 - 27) feet

= 8 feet

Now, there forms a right angled triangle, where hypotenuse is the distance that Ezra's bean bad has to cover.

Using Pythagoras theorem,

H^{2} = B^{2} + A^{2}

where H = hypotenuse,

B= Base of the triangle

A = Altitude of the triangle

we are given,

B = 15 feet,

A = 8 feet

On substituting the values,

H^{2} = 15^{2} + 8^{2}

H^{2} = 225 + 64\\H^{2} = 289\\H = \sqrt{289} \\H = 17\\

So hypotenuse = 17 feet

Ezra has to throw the beanbag 17 feet far in order to reach it to Allie.

To learn more about Pythagoras theorm from the given link

brainly.com/question/343682

#SPJ4

5 0
2 years ago
How to solve -2.4 × (-3.2)=
Zinaida [17]
The answer is 7.68.

Hope this helps.
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3 years ago
Read 2 more answers
Last question i will give brainliest
GarryVolchara [31]
H i think! sorry if it’s wrong
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3 years ago
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How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)
shepuryov [24]
LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x }  }  }{ \tan { x }  } \\ \\ =\frac { 1 }{ \tan { x }  } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x }  }  \right)

\\ \\ =\frac { \cos { x }  }{ \sin { x }  } \left( \frac { \sin ^{ 2 }{ x }  }{ \cos ^{ 2 }{ x }  } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x }  }  }{ \cos ^{ 2 }{ x }  }  \right) \\ \\ =\frac { \cos { x }  }{ \sin { x }  } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x }  }  }  }{ \cos ^{ 2 }{ x }  }  \right)

\\ \\ =\frac { \cos { x }  }{ \sin { x }  } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x }  \right)  }  }{ \cos ^{ 2 }{ x }  } \\ \\ =\frac { \cos { x }  }{ \sin { x }  } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x }  }  }{ \cos ^{ 2 }{ x }  } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x }  }{ \sin { x\cos ^{ 2 }{ x }  }  } \\ \\ =\frac { \sin ^{ 3 }{ x }  }{ \cos { x }  }

\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x }  }{ \cos { x }  } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x }  }{ \sin { x }  }  } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x }  } \\ \\ =\frac { \sin ^{ 2 }{ x }  }{ \cot { x }  } \\ \\ =RHS
8 0
2 years ago
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