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leonid [27]
3 years ago
6

Can someone help me on my last problem? #10

Mathematics
2 answers:
Mashutka [201]3 years ago
8 0
Ok but what did i do wrong and if you never studied the subject how did you know i was wrong
Kruka [31]3 years ago
8 0
Based on my small expertise in this subject,

the answer should be 1/(20)^22


the logic is the same as a coin flip
1st flip: call for heads, 1/2 chance
2nd: call for heads, (multiply by self) 1/2 x 1/2= 1/4 chance
3rd: heads, 1/4 x 1/2= 1/8
and so on...
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Find the value of x? please help​
Stella [2.4K]

Answer:

49

Step-by-step explanation:

With these types of problems, you have to subtract the outer and inner values and then divide by 2. So, (125-27)/2 = 49. Hope this helps!

4 0
3 years ago
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Joshua has a bag of marbles. In the bag are 5 white marbles, 3 blue marbles, and 7 green marbles. Peter randomly draws one marbl
Step2247 [10]

Answer:

1 / 5

Step-by-step explanation:

Given that:

Number of white marbles = 5

Number of blue marbles = 3

Number of green marbles = 7

Required is the approximate probability of drawing 2 green marbles, Note that drawing is done without replacement :

Probability = required outcome / Total possible outcomes

Total possible outcomes = sum of all marbles = (5 + 3 + 7) = 15 marbles

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P(Green) = 7 / 15

Second draw:

Required outcome = 7 - 1 = 6

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P(green) = 6 / 14

Probability of drawing out two green marbles :

(7/15 * 6/14) = 42 / 210 = 1 / 5

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3 years ago
Evaluate the expression when m = 4 and n = 6.<br><br> (3m−n)^2+4m
alexira [117]

Answer:

yor answer is 52

Step-by-step explanation:

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What is equal to 6 3/5
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