A local business had various advertising campaigns. It had television ads numbered 1,2,3,4,5,6, newspaper and magazine ads numbered 1,2,3,4,5, and social media ads labeled 1,2. If a single advertisement campaign is picked at random, what is the probability that the advertisement is a social media ad AND has an odd number? Provide the final answer as a fraction.

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

4

step-by-step explanation:

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation: