Answer:
heat required in pan B is more than pan A
Explanation:
Heat required to raise the temperature of the substance is given by the formula
now we know that both pan contains same volume of water while the mass of pan is different
So here heat required to raise the temperature of water in Pan A is given as
Now similarly for other pan we have
So here by comparing the two equations we can say that heat required in pan B is more than pan A
Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The is the overall medium,
is the medium of the first half, and
is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between and
. In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 + ) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
