The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
Answer:
00:13 mm:ss
Step-by-step explanation:
There are 60 seconds in a minute. This fact can be used to convert the time period(s) to minutes and seconds either before or after you do the subtraction.
<h3>Difference</h3>
It is often convenient to do arithmetic with all of the numbers having the same units. Here, we are given two values in seconds and asked for their difference.
100 s - 87 s = (100 -87) s = 13 s
The difference between the two time periods is 0 minutes and 13 seconds.
<h3>Conversion</h3>
If you like, the numbers can be converted to minutes and seconds before the subtraction. Since there are 60 seconds in a minute, the number of minutes is found by dividing seconds by 60. The remainder is the number of seconds that will be added to the time in minutes:
87 seconds = ⌊87/60⌋ minutes + (87 mod 60) seconds
= 1 minute 27 seconds
100 seconds = ⌊100/60⌋ minutes + (100 mod 60) seconds
= 1 minute 40 seconds
Then the difference is found in the same way we would find a difference involving different variables. (A unit can be treated as though it were a variable.)
(1 min 40 s) -(1 min 27 s) = (1 -1 min) + (40-27 s) = 0 min 13 s
The difference between the two time periods is 0 minutes and 13 seconds.
C - 0.15c is the same as 1c - 0.15c.
1 - 0.15 = 0.85.
Joe can also use 0.85c.
Given that <span>the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.
The probability of a normally distributed data between two values (a, b) is given by:

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