Someone else is answering the graph but the equation is y=-4x+2 -4 = slope 2 = y-intercept
Answer:
<h3><u>Mean</u></h3>
<u />



<h3><u>Standard Deviation</u></h3>



<h3><u>Summary</u></h3>
Nilo has a mean score of 10 and a standard deviation of 5.23.
Lisa has a mean score of 10 and a standard deviation of 1.83.
The <u>mean</u> scores are the <u>same</u>.
Nilo's standard deviation is higher than Lisa's. Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.
Answer:
The possible parking lengths are 45.96 feet and 174.031 feet
Step-by-step explanation:
Let x be the length of rectangular plot and y be the breadth of rectangular plot
A rectangular parking lot must have a perimeter of 440 feet
Perimeter of rectangular plot =2(l+b)=2(x+y)=440
2(x+y)=440
x+y=220
y=220-x
We are also given that an area of at least 8000 square feet.
So, 
So,

So,
General quadratic equation : 
Formula : 

So, The possible parking lengths are 45.96 feet and 174.031 feet