Since you always round up when you estimate in the middle lets say theres 420 tickets. Each is 7.95, but we'll round it to 8.00. 420x8.=3,360. After that part we need to find about how much tickets were refunded. So we round 18 to 20 and each ticket is still 8$. 20x8=160. Finally in all, we do 3,360-160=3200. Its an overestimate because we rounded up. Im not sure what else youre asking but thats the basic answer.

6 0

3 years ago

Which values are solutions to the inequality below?check all that apply x^2 >49 A. 6 B.20 C.-7 D.-5 E.-16 F. 7

MrRissso [65]

Well the square root of 49 is 7 so anything with an absolute value over 7 would work and if it is greater than or equal to, than 7 and -7 also work :)

4 0

4 years ago

I. There is a gap in the Potato Chips group

topjm [15]

Answer:

C) I and II

There is a gap in the Potato Chips group for the stem "14" and the Tortilla Chips group for the stem "12"

4 0

3 years ago

In how many ways can the 19 members of a chess club fill the offices of King, Knight, Bishop, and Rook?

goldfiish [28.3K]

Solution:

This problem is a permutation because the order matters here. This means that choosing A as King, B as Knight, C as Bishop and D as Rook results in a different arrangement from B as King, A as Knight, D as Bishop and C as Rook. We would count them both because in the first case A is King, but in the second case A is Knight.

Therefore, the possible number of ways are given below:

$19P4=\frac{19!}{(19-4)!} =\frac{19 \times 18 \times 17 \times 16 \times 15!}{15!} =19 \times 18 \times 17 \times 16 =93024$

Hence there will be 93024 ways 19 members of a chess club fill the offices of King, Knight, Bishop, and Rook.

3 0

3 years ago