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anonymous
 5 years ago
Somebody please help me with probability
anonymous
 5 years ago
Somebody please help me with probability

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all these are the ones i need help with

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the first pic.... 1. should be 7/20 which is 35% in otherwords D. You get the 20 from counting the lines from A to E which is 20. Then count the lines between BD which is 916 or 7. So you get 7/20. You basically do that for all of them...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for picture 38. Since a circle is 360 degrees and since you know each quarter should be 90 degrees. They give you 65 degrees so 6590 is 25 and you have 4 shaded 25s so you take the 25 degrees and multiply it by 4 to get 100. Then just take 100/360 and you should get 27.8%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so in the first picture for number 2, would it be .27?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would i do picture 37, 39, and 41?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a few minutes.. i know for 37 its above 50% for sure..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0On picture 41, all you have to do is imaging a dart board, just flip it to where the "Double points" is on the inside of the smallest circle. It's obviously answer choice C.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 37 if you move all the shade from one circle to the other you will have one complete circle... so that leaves you with one circle with no shade.. so you simply find the area of that by using pi(radius)^2 your radius is 4. Your rectangle area is 8x4 which is 32. so take the 32(unshaded circle area) = a number. then take that number and divide by 32 which gives you 64%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on its 60.7% i mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so picture 39 would 10.68%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not sure about 39 and 41.... sorry about that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay.. thank you for your help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0abigail, can u help me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya post a question up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Calvin caught five fish that had a mean of 19.3cm. Four of his fish measured: 17.5, 23, 15, and 23cm. Which of the following could be the length of the fifth fish that Calvin caught? A)15cm B)16cm C)17cm D)18cm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you basically just have to guess and check.. simply add those 4 numbers together and you can add any one of those (AD) to them then just divide it by 5 because you have 5 total fish. Mean = average so your taking the average of the fish you caught. Just continue until you get 19.3cm. Your 5th fish should be 18cm to maintain the 19.3 cm average

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i've got 16cm? What did i do wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats wrong.. 16+23+17.5+23+15 = 94.5/5 = 18.9cm which doesnt equal 19.3cm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok... so 23+23+17.5+15+18=96.5/5=19.3!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you help me on an open ended question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Write a set of data with at least five values and the given measures: a) a mean of five b) a mode of five c)a median of five d)a range of five
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