Answer: The total cost was $13,585.00.
Step-by-step explanation:
13 * 425 = 5,525
13 * 620 = 8,060
8,060 + 5,525 = 13,585
17 books in total, and 8 new. To find the number of used books, we subtract 8 from 17.
17-8= 9 used books
Now we have 8 new books and 9 used books.
All books: 17
Used books: 9
New books: 8
(a) all books: used books
17: 9
(b) new books: used books
8: 9
Answer:
Q1 - 2
Q1 - 5
Q3 - 7
maximum 9
minimum 1
Step-by-step explanation:
order the numbers from least to greatest
Step 1: We make the assumption that 498 is 100% since it is our output value.
Step 2: We next represent the value we seek with $x$x.
Step 3: From step 1, it follows that $100\%=498$100%=498.
Step 4: In the same vein, $x\%=4$x%=4.
Step 5: This gives us a pair of simple equations:
$100\%=498(1)$100%=498(1).
$x\%=4(2)$x%=4(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
$\frac{100\%}{x\%}=\frac{498}{4}$
100%
x%=
498
4
Step 7: Taking the inverse (or reciprocal) of both sides yields
$\frac{x\%}{100\%}=\frac{4}{498}$
x%
100%=
4
498
$\Rightarrow x=0.8\%$⇒x=0.8%
Therefore, $4$4 is $0.8\%$0.8% of $498$498.
1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
