A city pool sells full-day and half-day passes during the summer. The goal is to make $1000 each day from pool pass sales. The graph shows the number of full-day and half-day passes they need to sell to make $1000.

What is the maximum number of half-day passes the pool can sell and make exactly $1000?

The answer is 333 attending (to equal x=2) based on 666 in total attending. Then y=4 will equal the maximum per 6 visitors after counting any less or same amount that attended. We search for 1000 revenue and if the price was £1 for half and £2 for all day we would have maximum 1/2 based on peak days and 1/3 revenue.  The price therefore would be based on total numbers attending and using 1/3 to estimate the price then dividing this into 1000 to set the new price or multiplying the actual numbers of £1 by 1000 and expecting 2000 attendance. Half this attendance every time prices go up by £1.

Step-by-step explanation:

You can find a ratio if the prices are exactly half price = 1:2 thereafter for maximum, foresee that half price also has its own preference at a 1:2 ratio so this becomes 1:2 +1:2 =2:4 The ratio of 1000 to 2:4 along the axis line. Where 2 will always be half in monetary value it will also be half than its visitors. Even though it will be half the opted average 1:2 ratio. The income from half day projections turns to 1/3.

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.

step-by-step explanation: