50km to 80km is 30km
10c to -80c is -90c
and since it linear (Which means the change is constant or the same)
every 10 km show a -30c decrease
so
60 km is -20c
70km is -50c
answer is c
Answer:
Step-by-step explanation:
a) False. There are three factor pairs for 36:
1×36
2×18
3×12
4×9
6×6
b) True.
18 = 2×3²
Odd divisors of 18: 1, 3, 9
1+3+9 = 13
c) False
100 = 2×5²
25 = 5²
Greatest common divisor = 25
Answer:
The true statements are:
B. Interquartile ranges are not significantly impacted by outliers
C. Lower and upper quartiles are needed to find the interquartile range
E. The data values should be listed in order before trying to find the interquartile range
Step-by-step explanation:
The interquartile range is the difference between the first and third quartiles
Steps to find the interquartile range:
- Put the numbers in order
- Find the median Place parentheses around the numbers before and after the median
- Find Q1 and Q3 which are the medians of the data before and after the median of all data
- Subtract Q1 from Q3 to find the interquartile range
The interquartile range is not sensitive to outliers
Now let us find the true statements
A. Subtract the lowest and highest values to find the interquartile range ⇒ NOT true (<em>because the interquartial range is the difference between the lower and upper quartiles</em>)
B. Interquartile ranges are not significantly impacted by outliers ⇒ True <em>(because it does not depends on the smallest and largest data)</em>
<em />
C. Lower and upper quartiles are needed to find the interquartile range ⇒ True <em>(because IQR = Q3 - Q2)</em>
<em />
D. A small interquartile range means the data is spread far away from the median ⇒ NOT true (<em>because a small interquartile means data is not spread far away from the median</em>)
E. The data values should be listed in order before trying to find the interquartile range ⇒ True <em>(because we can find the interquartial range by finding the values of the upper and lower quartiles)</em>
Answer:
76,050 ft²
Step-by-step explanation:
If the area must be rectangular, let L be the length of the side opposite to the creek, and S be the length of the remaining two sides.
The perimeter of the fencing and the area of the pasture are:
The value of S for which the derivate of the area function is zero is the length of S that maximizes the area of pasture:
The maximum possible area is:
