Answer:
y-3
Problem:
What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?
Step-by-step explanation:
Dividend=quotient×divisor+remainder
So we have
xy-3=(x-1)×(y)+remainder
xy-3=(xy-y)+remainder *distributive property
Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.
We need to get rid of minus y so we need plus y in the remainder.
We also need minus 3 in the remainder.
So the remainder is y-3.
Let's try it out:
xy-3=(xy-y)+remainder
xy-3=(xy-y)+(y-3)
xy-3=xy-3 is what we wanted so we are done here.
The answer to you question is c
Order.
9, 12, 12, 14, 15, 16, 18, 21
Now split into quarters.
9, 12, 12, 14, 15, 16, 18, 21
| | |
(1) (3)
Determine the values of (1) and (3) by using medians.
14 + 12 + 12 + 9 / 4
47 / 4
approx. 12
So Q1 = 12.
15 + 16 + 18 + 21 / 4
70 / 4
approx. 17
Therefore the answer is C I think. I have not done this ever before. All the knowledge I did was from research lol
Thirty nine thousand and five
