10. The probability of buying pizza for dinner is 34% and the probability of buying a new car is 15%. The probability of buying a new car given that you eat pizza for
dinner is 42%. What is the probability of eating pizza for dinner given that they
buy a new car?

The probability of eating pizza given that a new car is bought is 0.952

Step-by-step explanation:

This kind of problem can be solved using Baye’s theorem of conditional probability.

Let A be the event of eating pizza( same as buying pizza)

while B is the event of buying a new car

P(A) = 34% = 0.34

P(B) = 15% = 15/100 = 0.15

P(B|A) = 42% = 0.42

P(B|A) = P(BnA)/P(A)

0.42 = P(BnA)/0.34

P(B n A) = 0.34 * 0.42 = 0.1428

Now, we want to calculate P(A|B)

Mathematically;

P(A|B = P(A n B)/P(B)

Kindly know that P(A n B) = P(B n A) = 0.1428

So P(A|B) = 0.1428/0.15

P(A|B) = 0.952

i think is g(1)=-1.

hope that i can

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2