Suppose U1 and U2 are i. i.d. Unif(0,1) withU1=0.1 and U2=0.8. Use the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variate. Don't forget to use radians instead of degrees. a. 0.326
b. 0.326
c. 0.663
d. 1.96

0.663 ( c )

Step-by-step explanation:

U1 = 0.1 , U2 = 0.8

using the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variable

first step : generate single obsⁿ from N ( -1,4 )

attached below is the detailed solution

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )