<h3>
Answer: 1/8</h3>
In decimal form, 1/8 = 0.125 which converts to 12.5%
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Work Shown:
The 63 should be 6^3. There are 6 choices per slot, and 3 slots, so 6^3 = 216 different outcomes.
Here are all of the ways to add to 11 if we had 3 dice
- sum = 1+4+6 = 11
- sum = 1+5+5 = 11
- sum = 1+6+4 = 11
- sum = 2+3+6 = 11
- sum = 2+4+5 = 11
- sum = 2+5+4 = 11
- sum = 2+6+3 = 11
- sum = 3+2+6 = 11
- sum = 3+3+5 = 11
- sum = 3+4+4 = 11
- sum = 3+5+3 = 11
- sum = 3+6+2 = 11
- sum = 4+1+6 = 11
- sum = 4+2+5 = 11
- sum = 4+3+4 = 11
- sum = 4+4+3 = 11
- sum = 4+5+2 = 11
- sum = 4+6+1 = 11
- sum = 5+1+5 = 11
- sum = 5+2+4 = 11
- sum = 5+3+3 = 11
- sum = 5+4+2 = 11
- sum = 5+5+1 = 11
- sum = 6+1+4 = 11
- sum = 6+2+3 = 11
- sum = 6+3+2 = 11
- sum = 6+4+1 = 11
There are 27 ways to add to 11 using 3 dice. This is out of 216 total outcomes of 3 dice being rolled.
So, 27/216 = (1*27)/(8*27) = 1/8 is the probability of getting 3 dice to add to 11.
Answer: 35
Step-by-step explanation:
Answer:
Yes, we can assume that the percent of female athletes graduating from the University of Colorado is less than 67%.
Step-by-step explanation:
We need to find p-value first:
z statistic = (p⁻ - p0) / √[p0 x (1 - p0) / n]
p⁻ = X / n = 21 / 38 = 0.5526316
the alternate hypothesis states that p-value must be under the normal curve, i.e. the percent of female athletes graduating remains at 67%
H1: p < 0.67
z = (0.5526316 - 0.67) / √[0.67 x (1 - 0.67) / 38] = -0.1173684 / 0.076278575
z = -1.538681
using a p-value calculator for z = -1.538681, confidence level of 5%
p-value = .062024, not significant
Since p-value is not significant, we must reject the alternate hypothesis and retain the null hypothesis.
Divide 4,320 between 9 and you'll get 408 per week. divide 408 between 40 and you'll get how much he makes per hour
