Is 1/3 bigger than 5/6

Answers

Answer 1

Answer: no because in 1/3 you get 2/6 so if you have 1/3 vs. 5/6 than you only have 1/3 containing 2/6 and we are looking for 5/6 so that is wrong. 5/6 is bigger that 1/3.

Answer 2

Answer: Make them both into 6ths. To do that, multiply 1/3 by 2/2.
1/3x2/2=2/6
No, 5/6 is bigger than 1/3 because it is basically only 2/6.
Hope this helped!

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If 120 percent is £360 what is 100%

Answers

To get from 120% to 100%, you simply divide by 1,2. 360 / 1,2 = 300

What is 8/11 in lowest terms.

Answers

8/11 9/10 are some examples of lowest terms
LowestTerm is a fraction that has the lowest termed. You will say if thefraction is in lowest term if the two has no common factor. Hope Ihelped.

It is already reduced to lowest terms because 11 cannot be reduced evenly but 8 can. So it is in its lowest terms.

Solve each inequality

3(x+6)-2(x+2)>10
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Answers

Solve:

3(x+6)-2(x+2)>10

3x+18-2x-4>10

x+14>10

The solution: x>-4

I hope I helped you with your solution! If you could give me brainliest, I would be very thankful!

How to do 200 + 1 and 1/2 x equals 548 - 1 and 2/5 x

Answers

So to do this you have to first calculate the sum subtract, multiply both sides by 10, move the terms, and collect like terms. So X=3460 I believe

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A local flower shop is selling $1 raffle tickets and is donating all proceeds to the children's hospital. They sold 5,500 tickets and plan to give away 20 flower arrangements valued at $70 and 20 gift certificates valued at $25. Find the expected value of purchasing a raffle ticket.Round to the nearest cent. Do not round until the final calculation.

Answers

The expected value of purchasing a raffle ticket is -0.647.

What is the Expected Value of a Probability & Statistical Analysis?

The expected value is derived in probability analysis by the multiplication of each of the potential possibilities by the likelihood that each result will occur and then adding all of those values together.

Using the formula below, we can calculate the expected value as follows:

$\mathbf{ E(X)= \sum^(n)_(i=1)X_i P(X_i)}$