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.
Answer:
(82/3, -111/2) or x=82/3, y=-111/2
Step-by-step explanation:
Answer:
Benchmark fractions: 4/12 5/12 6/12 7/12
Mixed Fractions: 1 1/2 2 3/4 5 1/4 8 2/8
Step-by-step explanation:
A mixed fraction is simply a fraction with a whole number.
A benchmark fraction is a fraction used to compare and order other fractions.
5 or more add one more
4 or less could care less(aka don't change the number)
93 right and round it to the nearest hundreds
So ig 93 than you look at the tens is it more than 5 or less than 4?
It's more than 5 so you add another number on the hundreds since there is nothing on the hundred place it is 100(Answer)if it was less than 4 it would just be 0. Hope I helped
Using the binomial distribution, it is found that there is a:
a) 0.9298 = 92.98% probability that at least 8 of them passed.
b) 0.0001 = 0.01% probability that fewer than 5 passed.
For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.
<h3>What is the binomial probability distribution formula?</h3>
The formula is:


The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 90% of the students passed, hence
.
- The professor randomly selected 10 exams, hence
.
Item a:
The probability is:

In which:




Then:

0.9298 = 92.98% probability that at least 8 of them passed.
Item b:
The probability is:

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

Hence:
0.0001 = 0.01% probability that fewer than 5 passed.
You can learn more about the the binomial distribution at brainly.com/question/24863377