Now as people entered the gym for the basketball tournament, members of the pep club were selling raffle tickets. they had $1.50 tickets and $5 tickets. the number of $1.50 tickets sold was two less than four times the number of $5 tickets sold, and the pep club raised $1,152 from the ticket sales. let x represent the number of $1.50 tickets sold and let y represent the number of $5 tickets sold. check all the correct answers

number of tickets of $5 sold = 418 and number of tickets of $1.5 sold = 105

Step-by-step explanation:

Let y be the number of $5 tickets sold and x be the number of $1.50 tickets sold.

Now from the statement " Number of $1.50 tickets sold was 2 less than the four times the number of $5 tickets sold".

Equation will be x = 4y - 2

4y - x = 2 (1)

For second equation statement is " The pep club raised $1152 from the ticket sales".

Equation will be 5y + (1.50)x = 1152 (2)

Now we multiply equation (1) by 1.5 and add it to equation (2)

6y - 1.5x + (5y + 1.5x) = 3 + 1152

11y = 1155

y = 105

Now we put y = 105 in equation (1)

x = 4×105 - 2

x = 420 -2

x = 418

Therefore, number of tickets of $5 sold = 418 and number of tickets of $1.5 sold = 105

a orange at the grocery store idk what other product that could be

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.