Answer:
I am not sure, ig it's D...
25.59- 9.99= 15.60
15.60/ 0.05 = 312
312 minutes this month
Answer:
no
Step-by-step explanation:
no they are not they are hard
Step-by-step explanation:
There are 40 DuBois students going on a field trip.
Each one pays their teacher $8.75 to cover admission to the museum and lunch.
Admission for the students is $120 and each one gets an equal amount to spend on lunch.
To Find :
How much will each DuBois student be able to spend on lunch.
Solution :
Amount they paid for admission :
\begin{gathered}P =\$ \dfrac{120}{40}\\\\P=\$3\end{gathered}P=$40120P=$3
Now, amount of money to spend on lunch is 8.75 -8.75−3 = $5.75 .
Hence, this is the required solution.
Answer:
$8.00
Step-by-step explanation:
The problem statement gives two relations between the prices of two kinds of tickets. These can be used to write a system of equations for the ticket prices.
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<h3>setup</h3>
Let 'a' and 'c' represent the prices of adult and children's tickets, respectively. The given relations can be expressed as ...
a - c = 1.50 . . . . . . . adult tickets are $1.50 more
175a +325c = 3512.5 . . . . . total revenue from ticket sales.
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<h3>solution</h3>
We are only interested in the price of an adult ticket, so we can eliminate c to give one equation we can solve for 'a'. Using the first equation, an expression for c is ...
c = a -1.50
Substituting that into the second equation, we have ...
175a +325(a -1.50) = 3512.50
500a -487.50 = 3512.50 . . . . . . simplify
500a = 4000 . . . . . . add 487.50
a = 8 . . . . . . . . . divide by 500
An adult ticket costs $8.
