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777dan777 [17]
3 years ago
14

Sally has a box that has 30 red balloons and 60 total balloons. What is the probability a randomly selected one is red?

Mathematics
1 answer:
Andreyy893 years ago
6 0
1/2 probability because 30 is half of 60.
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-75/8 -71/88 which is greater
Setler79 [48]
Trick question. See these negative signs? The left number is -9.375, and the right number is -0.8068181818181818...  Notice that even though the right number looks like a smaller answer, it is closer to 0 on a number line, because it is negative. -71/88 is the correct answer.

4 0
3 years ago
The octal system uses base 8. the only available digits would be 0, 1, 2, 3, 4, 5, 6, and 7. if the value 2468 is converted to d
marysya [2.9K]
Your description and expansion suggest you want to evaluate
.. 246₈ = 2*8² +4*8¹ +6*8⁰
.. = 2*64 +4*8 +6
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5 0
3 years ago
1. Perform the indicated operations, and then answer the question.
Alexandra [31]
A is the correct answer
4 0
2 years ago
Please help due today
FromTheMoon [43]

Answer:

It's gotta be either A or B because C and D don't even make up a square.

I believe that it's one of those ( A or B).

Step-by-step explanation:

Sorry I couldn't help you all the way through to the answer. Hope this helps though! :)

8 0
2 years ago
Sunhee had four plastic shapes a square, a circle, and a pentagon in how many ways can she line up the four shapes of the circle
julsineya [31]

Answer:

only three shapes are mentioned.

But anyhow, 4 shapes allow the following reasoning...

you can start lining up with any of the for shapes, so you have 4 possibilities, for the second position you have 3 possibilities, for the third 2 and for the last only 1.

so 4*3*2*1= 24 possible ways

but with the added rule to keep the circle and the square apart, we have to reduce the number.

by the same step-by-step reasoning we get

4 possibilities for the 1st item

only 2 possibilities for the 2nd item

(if 1st is square or circle, the other one's not an option, if it's not we have to choose one of them to not place them both at point 3 and 4)

only 1 choice for the 3rd item

(reasoning as above, if 2nd is square or circle, we have to the only other, if it isn't, we can put it here at 3rd position)

plus scenarios where either square or circle are on 1st point, the other's on 4th.

I guess that makes 14 possible arrangements

5 0
3 years ago
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