Answer:
0%
Step-by-step explanation:
ITS A CUBE BRUH!!!!
Answer:
Price = $4.50
Maximum revenue = $2,430
Step-by-step explanation:
The concession stand sells 600 bags of peanut at $4.00 (600, $4.00).
If there is a $1 increase in price, the number of bags old will decrease by 120, which mens that 480 bags of peanuts will be sold at $5.00 (480, $5.00).
Tracing a linear relationship between price and quantity sold with the two given points:
![m = \frac{5-4}{480-600}\\ m=-\frac{1}{120}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B5-4%7D%7B480-600%7D%5C%5C%20m%3D-%5Cfrac%7B1%7D%7B120%7D)
![P - P_0=m*(Q-Q_0)\\P-4.00=-\frac{1}{120} (Q-600)\\P= -\frac{1}{120}Q +9](https://tex.z-dn.net/?f=P%20-%20P_0%3Dm%2A%28Q-Q_0%29%5C%5CP-4.00%3D-%5Cfrac%7B1%7D%7B120%7D%20%28Q-600%29%5C%5CP%3D%20-%5Cfrac%7B1%7D%7B120%7DQ%20%2B9)
The revenue function is given by the price multiplied by the quantity sold:
![R=Q*P= (-\frac{1}{120}Q +9)*Q\\R=-\frac{1}{120}Q^2 +9Q](https://tex.z-dn.net/?f=R%3DQ%2AP%3D%20%28-%5Cfrac%7B1%7D%7B120%7DQ%20%2B9%29%2AQ%5C%5CR%3D-%5Cfrac%7B1%7D%7B120%7DQ%5E2%20%2B9Q)
The value of 'Q' for which the derivate of the revenue function is zero, is the output level for which revenue is maximum:
![R'=0=-\frac{2}{120}Q+9\\Q_{max} = 540\\P_{max} = -\frac{540}{120} +9\\P_{max} =\$4.50](https://tex.z-dn.net/?f=R%27%3D0%3D-%5Cfrac%7B2%7D%7B120%7DQ%2B9%5C%5CQ_%7Bmax%7D%20%3D%20540%5C%5CP_%7Bmax%7D%20%3D%20-%5Cfrac%7B540%7D%7B120%7D%20%2B9%5C%5CP_%7Bmax%7D%20%3D%5C%244.50)
The total revenue of 540 units at $4.50 per unit is:
![R = \$4.50*540\\R=\$2,430](https://tex.z-dn.net/?f=R%20%3D%20%5C%244.50%2A540%5C%5CR%3D%5C%242%2C430)
Answer:
The 95% confidence interval for the population mean is between $140.89 and $159.11.
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 the margin of error 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{36}{\sqrt{150}} = 9.11](https://tex.z-dn.net/?f=M%20%3D%201.96%5Cfrac%7B36%7D%7B%5Csqrt%7B150%7D%7D%20%3D%209.11)
The lower end of the interval is the sample mean subtracted by M. So it is 150 - 9.11 = $140.89
The upper end of the interval is the sample mean added to M. So it is 150 + 9.11 = $159.11
The 95% confidence interval for the population mean is between $140.89 and $159.11.
228 - 48 = 180
180/48 = 3.75
Its increase 375%
Answer:
The correct answers are the second and third ones.
a=6,b=8,c=10
a=5,b=6,c=61 sqred
Step-by-step explanation:
for the second one, 36+64=100. So 100sqrd is 10.
For the third one, 25+36=61. So 61sqrd.