Answer:
B: 25%
Step-by-step explanation:
$90.00 - $67.50 = $22.5
$22.5 / $90 = 0.25 = 25%
Answer:
B: 25%
Answer:4 in each package
Step-by-step explanation:
Answer:
63 Hope this helps you and everyone else :)
Step-by-step explanation:
First, multiply 9x2 to get 18. Then, multiply 9x5 to get 45. Finally add both numbers to get 63.
The following are the ages of 13 history teachers in a school district. 24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56 Notic
pishuonlain [190]
The five-number summary and the interquartile range for the data set are given as follows:
- Interquartile range: 50 - 29 = 21.
<h3>What are the median and the quartiles of a data-set?</h3>
- The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
- The first quartile is the median of the first half of the data-set.
- The third quartile is the median of the second half of the data-set.
- The interquartile range is the difference between the third quartile and the first quartile.
In this problem, we have that:
- The minimum value is the smallest value, of 24.
- The maximum value is the smallest value, of 56.
- Since the data-set has odd cardinality, the median is the middle element, that is, the 7th element, as (13 + 1)/2 = 7, hence the median is of 43.
- The first quartile is the median of the six elements of the first half, that is, the mean of the third and fourth elements, mean of 29 and 29, hence 29.
- The third quartile is the median of the six elements of the second half, that is, the mean of the third and fourth elements of the second half, mean of 49 and 51, hence 50.
- The interquartile range is of 50 - 29 = 21.
More can be learned about five number summaries at brainly.com/question/17110151
#SPJ1
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
