The outlier, 78, is removed from the data set shown. 23, 23, 25, 27, 34, 34, 35, 41, 45, 45, 78 How does this affect the range?
KATRIN_1 [288]
Hello.
The range with 78 is 55
without 78 is 22
So the range goes down
Have a nice day
The answer to ur question is a
The graph of the Evan's data is attached below.
Dylan noted the time of day and the temperature, in degrees Fahrenheit, and his findings are as follows: At 6 a.m., the temperature was 58° F. For the next 3 hours, the temperature rose 1° per hour. For the next 4 hours, it rose 2° per hour. The temperature then stayed steady until 6 p.m. For the next 3 hours, the temperature dropped 2° per hour. The temperature then dropped steadily until the temperature was 62° at midnight. The data points are in the form of time and temperature. The data points are 6 AM = 58, 7 AM = 59, 8 AM = 60, 9 AM = 61, 10 AM = 63, 11 AM = 65, 12 PM = 67, 1 PM = 69, 6 PM = 69, 7 PM = 67, 8 PM = 65, 9 PM = 63, and 12 AM = 62.
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Answer:
A. a cube
B. a square prism
C. a rectangular prism
Step-by-step explanation:
The net of a given shape is the description of its three dimensional surfaces in two dimensional plane. From the given question, the 6 rectangular net of the shape has a condition that: the 6 rectangles may or may not be squares. Thus, the figure could be;
i. a cube
ii. a square
iii. a rectangular prism
The three options above have surfaces that could either be squares or rectangles.
9514 1404 393
Answer:
3a by 4a
Step-by-step explanation:
For dimensions L and W, the area and perimeter are ...
A = LW = 12a^2
P = 2(L+W) = 14a
Using the second equation, we can find L:
L +W = 7a . . . . . divide by 2
L = 7a -W
Substituting into the area formula gives the quadratic ...
(7a -W)(W) = 12a^2
W^2 -7aW +12a^2 = 0 . . . . arrange in standard form
(W -3a)(W -4a) = 0 . . . . . . . factor (find factors of 12 that total 7)
Then we have the two solutions ...
W = 3a, L = 4a
W = 4a, L = 3a
The rectangle dimensions are 3a by 4a.