I think the answer is around 1.4 as you need to multiply the area by 1/9 after finding the real area.
To find the interquartile range, you will list the data that is presented in the stem and leaf plot.
Find the median of the data (30.5)
Find the median of the lower half and the median of the upper half.
Subtract these two values.
The data are <u>20</u>, 25, 30, 30, 31, 40, 41, <u>49</u>.
27.5 40.5
40.5-27.5 = 13
The interquartile range is 13.
Answer:
Step-by-step explanation:
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by
T(x) = 160-0.05x^2
a. [0, 10]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
The average temperature
= (160 + 155)/2 = 157.5
b. [10, 40]
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (80 + 155)/2 = 117.5
c. [0, 40]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (160 + 80)/2 = 120
Answer:
the slope of a horizontal line is 0
Can you please add the graph, plus the question?
