An experiment involves tossing a pair of dice, one green and one red, and recording the numbers that come up.

(a) list the elements corresponding to the event a that the sum is greater than 7;
(b) list the elements corresponding to the event b that a number greater than 5 comes up on the green die;
(c) list the elements corresponding to the event a ∩ b

A ≡ { ( 2, 6 ), (  3, 5 ), ( 3, 6 ), ( 4, 4 ), (  4, 5 ), ( 4, 6 ), ( 5, 3 ), ( 5, 4 ), (  5, 5 ), ( 5, 6 ), ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

B ≡ { ( 6, 1 ), ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

A ∩ B ≡ { ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

Step-by-step explanation:

Tossing a pair of dice, one green and one red we will have 36 number of outcomes that is below

Let the first outcome be of green and second outcome be of red.

{ ( 1, 1 ), ( 1, 2 ), ( 1, 3 ), ( 1, 4 ), (  1, 5 ), ( 1, 6 )

 ( 2, 1 ), ( 2, 2 ), ( 2, 3 ), ( 2, 4 ), ( 2, 5 ), ( 2, 6 )

 ( 3, 1 ), ( 3, 2 ), ( 3, 3 ), ( 3, 4 ), (  3, 5 ), ( 3, 6 )

 ( 4, 1 ), ( 4, 2 ), ( 4, 3 ), ( 4, 4 ), (  4, 5 ), ( 4, 6 )

 ( 5, 1 ), ( 5, 2 ), ( 5, 3 ), ( 5, 4 ), (  5, 5 ), ( 5, 6 )

 ( 6, 1 ), ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

Now,

a ) List the elements corresponding to the event A that the sum is greater than 7.

A ≡ { ( 2, 6 ), (  3, 5 ), ( 3, 6 ), ( 4, 4 ), (  4, 5 ), ( 4, 6 ), ( 5, 3 ), ( 5, 4 ), (  5, 5 ), ( 5, 6 ), ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

b ) List the elements corresponding to the event B that a number greater than 5 comes up on the green die.

B ≡ { ( 6, 1 ), ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

c ) List the elements corresponding to the event A ∩ B.

A ∩ B ≡ { ( 6, 2 ), ( 6, 3 ), ( 6, 4 ), (  6, 5 ), ( 6, 6 ) }

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

£56

step-by-step explanation: