Answer:
50.40
Step-by-step explanation:
7 hours and 20 minutes times 7= 50 hours and 40 minutes
And then you have to change it into a mixed number and it is 50.40
19 small mowers and 11 large mowers are sold
<em><u>Solution:</u></em>
Let "a" be the number of small mowers sold
Let "b" be the number of large mowers sold
Cost of each small mower = $ 249.99
Cost of each large mower = $ 329.99
<em><u>30 total mowers were sold</u></em>
Therefore,
a + b = 30
a = 30 - b ------------- eqn 1
<em><u>The total sales for a given year was $8379.70</u></em>
<em><u>Thus we frame a equation as:</u></em>
number of small mowers sold x Cost of each small mower + number of large mowers sold x Cost of each large mower = 8379.70

249.99a + 329.99b = 8379.70 ---------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
<em><u>Substitute eqn 1 in eqn 2</u></em>
249.99(30 - b) + 329.99b = 8379.70
7499.7 - 249.99b + 329.99b = 8379.70
80b = 8379.70 - 7499.7
80b = 880
Divide both sides by 80
b = 11
<em><u>Substitute b = 11 in eqn 1</u></em>
a = 30 - 11
a = 19
Thus 19 small mowers and 11 large mowers are sold
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE