Answer:
58.3
Step-by-step explanation:
58.3 has the highest unit (the number left of the dot)
for 6
in a
given
slope(m)=1
point=(2,-4)=(x1,y1)
we know
equation of straight line
y-y1=m(x-x1)
y+4=1(x-2)
y+4=x-2
therefore x-y-6=0 is the required equation
substituting with ax+by+c=0 we get
a=1
b= -1
c= -6
Answer:
The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
Now, find M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 1.96*\frac{17.4}{\sqrt{73}} = 3.99](https://tex.z-dn.net/?f=M%20%3D%201.96%2A%5Cfrac%7B17.4%7D%7B%5Csqrt%7B73%7D%7D%20%3D%203.99)
The lower end of the interval is the sample mean subtracted by M. So it is 230 - 3.99 = 226.01
The upper end of the interval is the sample mean added to M. So it is 230 + 3.99 = 233.99.
The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.
the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it? well 20% of "x" or namely 0.2x. Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.
![\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bsolution%7D%7Bquantity%7D%26%5Cstackrel%7B%5Ctextit%7B%5C%25%20of%20%7D%7D%7Bamount%7D%26%5Cstackrel%7B%5Ctextit%7Bliters%20of%20%7D%7D%7Bamount%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7B1st%20solution%7D%26x%260.20%260.2x%5C%5C%20%5Ctextit%7B2nd%20solution%7D%26y%260.60%260.6y%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20mixture%26100%260.44%2644%20%5Cend%7Barray%7D~%5Chfill%20%5Cbegin%7Bcases%7D%20x%2By%3D100%5C%5C%5C%5C%200.2x%2B0.6y%3D44%20%5Cend%7Bcases%7D)
![x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}](https://tex.z-dn.net/?f=x%2By%3D100%5Cimplies%20y%3D100-x~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B0.2x%2B0.6%28100-x%29%3D44%7D%20%5C%5C%5C%5C%5C%5C%200.2x%2B60-0.6x%3D44%5Cimplies%20-0.4x%2B60%3D44%5Cimplies%20-0.4x%3D-16%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B-16%7D%7B-0.4%7D%5Cimplies%20%5Cboxed%7Bx%3D40%7D~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B100-40%7D%7By%3D60%7D%7D)
about 1 million robux cuz they each wortha bout a cent idk though on black market is cheaper