The power increases by 21% per hour. 21% written as a decimal is 0.21.
Because it increases you would multiply the starting value by 1.21 times the number of hours (t).
That needs to be less than or equal to 100,00
The equation would be:
C: 15,040(1.21)t ≤ 100,000
It would help if we represent the information in form of two-ways table as shown below
The number of students who are not on probation is 104 students
There are 23 students who are not on probation and is satisfied is 81 students
The probability of students not on probation and is satisfied is 81/104
What? This isn’t a question
<u>Given</u>:
Given that the data are represented by the box plot.
We need to determine the range and interquartile range.
<u>Range:</u>
The range of the data is the difference between the highest and the lowest value in the given set of data.
From the box plot, the highest value is 30 and the lowest value is 15.
Thus, the range of the data is given by
Range = Highest value - Lowest value
Range = 30 - 15 = 15
Thus, the range of the data is 15.
<u>Interquartile range:</u>
The interquartile range is the difference between the ends of the box in the box plot.
Thus, the interquartile range is given by
Interquartile range = 27 - 18 = 9
Thus, the interquartile range is 9.
Answer:
9m^2+17m-9
Step-by-step explanation: