Answer:
The length and width of the parking lot is 78 meters and 114 meters respectively.
Step-by-step explanation:
Given;
Perimeter of the parking lot = 
Solution,
Let the width of the parking lot be x.
Then, according to question length = (x-36).
The perimeter of a rectangle is sum of all the sides of rectangle. Which is given by an expression;

Now substituting the values, we get;

Width = 
Length = 
Hence the length and width of the parking lot is 78 meters and 114 meters respectively.
All the angles in a circle equal to 360 degrees. So basically just add up all the known values: 23+23+15=61 and then subtract that from 360: 360-61=299.
You can also do 23+23+15+x=360. Adding it all becomes 61+x=360, subtract 61 from both sides of the equation, and you get x=299.
(Something to help remember is that the angle is obtuse, meaning it’s automatically more than 90 degrees, hope this helps!)
Answer:
noooooooo
Step-by-step explanation:
(x−a)(x−b)
=(x+−a)(x+−b)
=(x)(x)+(x)(−b)+(−a)(x)+(−a)(−b)
=x2−bx−ax+ab
<h2><u><em>
=ab−ax−bx+x2</em></u></h2>
Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case