well, the assumption is that is a rectangle, namely it has two equal pairs, so we can just find the length of one of the pairs to get the dimensions.
hmmmm let's say let's get the length of the segment at (-1,-3), (1,3) for its length
and
the length of the segment at (-1, -3), (-4, -2) for its width
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{length}{L}=\sqrt{[1-(-1)]^2+[3-(-3)]^2}\implies L=\sqrt{(1+1)^2+(3+3)^2} \\\\\\ L=\sqrt{4+36}\implies L=\sqrt{40} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Blength%7D%7BL%7D%3D%5Csqrt%7B%5B1-%28-1%29%5D%5E2%2B%5B3-%28-3%29%5D%5E2%7D%5Cimplies%20L%3D%5Csqrt%7B%281%2B1%29%5E2%2B%283%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20L%3D%5Csqrt%7B4%2B36%7D%5Cimplies%20L%3D%5Csqrt%7B40%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{width}{w}=\sqrt{[-4-(-1)]^2+[-2-(-3)]^2}\implies w=\sqrt{(-4+1)^2+(-2+3)^2} \\\\\\ w=\sqrt{9+1}\implies w=\sqrt{10} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{A=Lw}\implies \sqrt{40}\cdot \sqrt{10}\implies \sqrt{400}\implies \boxed{20}](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-4%7D~%2C~%5Cstackrel%7By_2%7D%7B-2%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bwidth%7D%7Bw%7D%3D%5Csqrt%7B%5B-4-%28-1%29%5D%5E2%2B%5B-2-%28-3%29%5D%5E2%7D%5Cimplies%20w%3D%5Csqrt%7B%28-4%2B1%29%5E2%2B%28-2%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20w%3D%5Csqrt%7B9%2B1%7D%5Cimplies%20w%3D%5Csqrt%7B10%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rectangle%7D%7D%7BA%3DLw%7D%5Cimplies%20%5Csqrt%7B40%7D%5Ccdot%20%5Csqrt%7B10%7D%5Cimplies%20%5Csqrt%7B400%7D%5Cimplies%20%5Cboxed%7B20%7D)
This is the question:
A
bicycle manufacturing company makes a particular type of bike.
Each
child bike requires 4 hours to build and 4 hours to test.
Each
adult bike requires 6 hours to build and 4 hours to test.
With
the number of workers, the company is able to have up to 120 hours of building
time and
100 hours of testing time for a week.
If
c represents child bikes and a represents adult bikes,
determine
which system of inequality best explains whether the company can build 10 child
bikes and 12 adult bikes in the week
Now you
can state the system of inequalities from the statements
1) First inequality based on the hours availble
to buiding
Each
child bike requires 4 hours, e<span>ach
adult bike requires 6 hours to build and </span>the company is able to have up to 120 hours of building =>
4c + 6a ≤ 120
2) Second inequality based of the hours available to testing.
Each
child bike requires 4 hours to test, each
adult bike 4 hours to test and the company is able to have up 100 hours of testing time for a week =>
4c + 4a ≤ 100
Then the two inequalities are:
4c + 6a ≤ 1204c + 4a ≤ 100<span>
The answer is Yes, because the bike order meets the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100Which you can verify by replacing in both equations 10 for c and 12 for a. Look:
1) 4(10) + 6(12) = 40 + 72 = 112 ≤ 1202) 4(10) + 4(12) = 40 + 48 = 88 ≤ 100</span>
The 90% , 99% confidence interval for the population mean is 32.145 < 
< 35.855 and 31.093 < < 36.907
<h3>What is Probability ?</h3>
Probability is the study of likeliness of an event to happen.
It is given that
Total Population = 50
Mean = 35
The confidence interval is given by
is the mean
z is the confidence level value
s is the standard deviation
n is the population width
(a) The 90% confidence interval for the population mean
90% 
= 0.05
Z = 1.64
34 
1.64 * 8 / √50
34 1.855
32.145 < < 35.855
(b) The 99% confidence interval for the population mean
99% = 0.005
Z=2.57
34 2.57 * 8 / √50
34 2.907
31.093 < < 36.907
Therefore the confidence interval for population mean has been determined.
The complete question is
A simple random sample of 50 items from a population width =7 resulted in a sample mean of 35. If required, round your answers to two decimal places.
a. Provide a 90% confidence interval for the population mean
b. Provide a 99% confidence interval for the population mean
To know more about Probability
brainly.com/question/11234923
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