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Readme [11.4K]
3 years ago
6

99 POINTS!!!Abel uses a probability simulator to roll a six-sided number cube 100 times and to flip a coin 100 times. The result

s of the experiment are shown below:
Number on the Cube
1 12
2 18
3 30
4 22
5 10
6 8


Heads Tails
34 66


Using Abel's simulation, what is the probability of rolling a 2 on the number cube and the coin landing on tails?

A)84/10,00... a isnt correct though
B)1,188/10,000
C)18/100
D)66/100
What the answer and how do i show the work????
Mathematics
2 answers:
igor_vitrenko [27]3 years ago
6 0
Well first of all, flipping a coin is not a 50/50 chance, it is a 49/51 chance favoring the side you started on. I would point out that the flipping of the coin seems to favor tails with 66 to 34, so it is reasonable to assume that the coin was flipped with tails side up more often than heads. First you need to add 66+34=100, the answer is B
Step2247 [10]3 years ago
5 0
The running of the simulation serves to give us an approximate probability of each of the outcomes, so in this case, P(Tails)=66/100 and P(2)=18/100.

The probability of two independent things occurring can be written as P(A)*P(B).

Therefore, the probability of a 2 and tails is (66/100)*(18/100)=0.1188
= 1188/10000
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side length only
area = 6 * 12^2 * / 4*tan(30)
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If apothem and side length were given with more precision, the answers would be closer.

Source:
http://www.1728.org/polygon.htm


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enot [183]

Answer:

$10 \frac{14}{15} $

Step-by-step explanation:

Given :

$\frac{21}{4}-\left(\frac{13}{5}+\frac{\frac{25}{6}}{\frac{25}{8}} \right)$

$= \frac{21}{4}-\left(\frac{13}{5}+\frac{25}{6} \times \frac{8}{25} \right)$

$=\left( \frac{21}{4}-\frac{13}{5}\right)+\frac{4}{3} $

$=\left( \frac{105-52}{20}\right)+\frac{4}{3} $

$= \frac{53}{20}+\frac{4}{3} $

$= \frac{53}{5}+\frac{1}{3} $

$=\frac{159+5}{15}$

$=\frac{164}{15}$

$=10 \frac{14}{15} $

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