For this case we must solve the following system of equations:
We multiply the first equation by -2:
We add the new equation with the second one:
We have different signs subtracted and the sign of the major is placed:
Now we find the value of the variable "y":
Thus, the solution of the system is given by:
Answer:
Answer:
A
Step-by-step explanation:
because it goes up constantly by 12
Answer:
B. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.94 units.
Step-by-step explanation:
First step is to determine the length of EF, since that will give us 2 sides of the triangle (since EG = EF).
From the diagram, we can easily make a rectangle triangle by dropping a vertical line from vertex E, let's name Z the meeting point of that line with the segment GF. Then we have a rectangle triangle EZF with a height of 4 and a base of 2, of which EF is the hypotenuse. So...
EF² = 4² + 2² = 16 + 4 = 20
EF = √20 = 4.47
Now that we have EF, we also have EG:
EF = 4.47
EG = 4.47
GF = 4 (visible on the graph)
Perimeter = 4.47 + 4.47 + 4 = 12.94 units.
Answer:
The sample size to obtain the desired margin of error is 160.
Step-by-step explanation:
The Margin of Error is given as
Rearranging this equation in terms of n gives
![n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2](https://tex.z-dn.net/?f=n%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2)
Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as
![n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n](https://tex.z-dn.net/?f=n_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM_2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%2F2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B2%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D2%5E2%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4n)
As n is given as 40 so the new sample size is given as
So the sample size to obtain the desired margin of error is 160.
You have to divide the total cost $12 by the 4 bars, so $12/4 bars = $3 per bar
