Answer:
a. The estimated coefficient for size is approximately <u>13.81</u>.
b. In the regression, two predictors are used. These two predictors are size and fireplace (FP).
Explanation:
a. The estimated coefficient for size is approximately _____.
Estimated coefficient for size = Standard Error of size * t-Stat of size = 1.2072436 * 11.439 = 13.81
Therefore, the estimated coefficient for size is approximately <u>13.81</u>.
b. How many predictors (independent variables) were used in the regression?
Independent variables can be described as variables that are changed or manipulated in order to measure the effect of their changes on the dependent variable. Independent variables are therefore also called predictors because they employed to predict the dependent variable.
In the regression, two predictors are used. These two predictors are size and fireplace (FP).
Answer: The Option "d.returning inventory that is defective or broken" is NOT an example of safeguarding inventory.
Explanation: If we analyze the statements:
a.physical devices such as two-way mirrors, cameras, and alarms - These are all tools intended for protection against possible inventory theft.
b.storing inventory in restricted areas - Restricting access only to inventory-enabled personnel is able to protect the inventory much more than if anyone can access it.
c.matching receiving documents, purchase orders, and vendor's invoice - Controlling each of the purchase documents and performing the physical count reduces the possibilities of inventory differences for losses or errors.
d.returning inventory that is defective or broken - Returning the defective inventory is a post-echo action that occurred due to the unprotection of the inventory, therefore it could not be referred to as an example of inventory protection.
Answer:
![STC = 20K + 25L = 20*5 + 25*[\frac{Q^2}{25}] = 100 + Q^2](https://tex.z-dn.net/?f=%20STC%20%3D%2020K%20%2B%2025L%20%3D%2020%2A5%20%2B%2025%2A%5B%5Cfrac%7BQ%5E2%7D%7B25%7D%5D%20%3D%20100%20%2B%20Q%5E2%20)
Explanation:
We are given:
K units of capital and L units of labor.
•Each unit of capital cost = 20
• Each unit of labor cost =25
• Level K is fixed at 5 units
We are told production function Q = K√L
Using the production functions and the values given, we can get that Q=5√L.
To find Q, the amount of labor will be given as:
Therefore, the Short run total cost function (STC) will be:
![20K + 25L = 20*5 + 25[\frac{Q^2}{25}] = 100 + Q^2](https://tex.z-dn.net/?f=%2020K%20%2B%2025L%20%3D%2020%2A5%20%2B%2025%5B%5Cfrac%7BQ%5E2%7D%7B25%7D%5D%20%3D%20100%20%2B%20Q%5E2%20)
Answer:
c. 24.78%
Explanation:
For computing the expected standard deviation first we have to find out the expected rate of return which is shown below:
Expected rate of return = Respective return × Respective probability
=(0.4 × -10) + (0.2 × 10) + (0.4 × 45)
= 16%
Now we have to find out the total probability which is shown below:
Probability Return Probability × (Return - Expected Return)^2
0.4 -10 0.4 × (-10-16)^2 = 270.4
0.2 10 0.2 × (10 - 16)^2 = 7.2
0.4 45 0.4 × (45 - 16)^2 = 336.4
Total = 614%
As we know that
So
Standard deviation= [Total probability × (Return - Expected Return)^2 ÷ Total probability]^(1 ÷2)
= (614)^(1 ÷ 2)
= 24.78%
