0.99 0.89 7/8 least to greatest

Plzz help asp need really smart peeps

seraphim [82]

Answer:

0.755

Step-by-step explanation:

The mixed fraction just needs to simplified to a decimal number

like so:

0.755

200/151

151-140 = 11

11-10=1

1-1=0

4 0

3 years ago

What inverse operation do you use to solve for m? <br><br> 11=m - 4

gulaghasi [49]

Answer:

I would use addition

Step-by-step explanation:

Since addition is the opposite of subtraction.

7 0

2 years ago

Read 2 more answers

An instructor gives her class the choice to do 7 questions out of the 10 on an exam.

Maksim231197 [3]

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

$10C7=\frac{10!}{7!(10-7)!}=120$

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

$(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50$

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

$(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50$

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

$(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10$

So, if a student must answer at least 3 of the first 5 questions, he/she have 110 choices. It is calculated as:

50 + 50 + 10 = 110

6 0

3 years ago

Which equation should you solve to find x?

devlian [24]

the answer is B ........

3 0

1 year ago

Do you think the equations (x−1)(x+3)=17+x and (x−1)(x+3)+500=517+x should have the same solution set? Why?

oksano4ka [1.4K]

Answer:

Those two pair of equations have the same solution set.

Step-by-step explanation:

There are two equations

(x-1)(x+3)=17+x ..... (1) and

(x-1)(x+3)+500=517+x ...... (2)

We have to check the same solution set will be there for equations (1) and (2) or not.

Now, we are going to rearrange the equation (2).

(x-1)(x+3)+500=517+x

⇒ (x-1)(x+3)=517-500+x

⇒(x-1)(x+3)=17+x

This is the same equation as equation (1).

Therefore, there will be the same solution set for equations (1) and (2). (Answer)

There are two equations

(x-1)(x+3)=17+x ..... (3) and

3(x-1)(x+3)+500=51+3x ...... (4)

We have to check the same solution set will be there for equations (3) and (4) or not.

Now, we are going to rearrange the equation (4).

3(x-1)(x+3)+500=51+3x

⇒ 3(x-1)(x+3)=3(17+x)

⇒(x-1)(x+3)=17+x

This is the same equation as equation (3).

Therefore, there will be the same solution set for equations (3) and (4). (Answer)

7 0

3 years ago