The experimental probability is computed to be 43/150 or approximately 28.67%. This is computed by dividing the event of number 3 showing by the total number of times the cube is rolled.
The theoretical probability is computed to be 1/6 or approximately 16.67%. Since there is only one side with the number 3, and there is a total of six sides in a cube. Theoretical probability assumes that the number cube is fair and all sides have equal chances of showing up.
Answer:
<h2><u><em>
the circumference of a circle is defined as:</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
c = pi*diameter</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
if we have x segments of length y, then we can fit:</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
c/y = z</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
we can fit z segments of size y on the circumference c</em></u></h2><h2><u><em>
</em></u></h2>
<u><em></em></u>
Answer:
1/2
Step-by-step explanation:
We first find the least common multiple, which is 6. Therefore, we can multiply 1/3 x 2/2 (since 2/2 is equal to one and won't change the final amount) and get 2/6. 2/6+1/6=3/6, or 1/2.
Answer:
This can be solved by using the empirical rule for a normal distribution.
Step-by-step explanation:
A. The number of skateboards given is one standard deviation above the mean. Approximately 68% of the data points lie within the range plus and minus one standard deviation of the mean. Therefore the required percentage is:
68 + 16 = 84%.
B. The given number of skateboards is two standard deviations above the mean. Approximately 95% of the data points lie within the range plus and minus two standard deviations of the mean. Therefore the required percentage is:
5/2 = 2.5%
C.The given number of skateboards is one standard deviations below the mean. Therefore the required percentage is:
16%.
z = -2.83
<u>Explanation:</u>
Given:
Mean, μ = 68
Standard deviation, σ = 6
Data point, x = 51
Z score, z = ?
We know:
Substituting the value we get:
A negative z score says that the data point is below average.
