The answer is 5x-5that is the answer
Answer:
Step-by-step explanation:
Given that a firm has a price of $5, an average total cost of $7, and an average variable cost of $4
Price = 5
Var cost = 4
Contribution = 1 dollar per unit
Since contribution is positive, there is scope for getting profit by increasing production.
In the short run, you should __operate______(operate/shut down) because __Price______exceeds ________ average variable cost price . In the long run, you should __exit______(stay in/exit) the market because ________ average total cost price exceeds____price.______average variable cost price average total cost
Answer:
probability of lasting longer = 1.7%
Step-by-step explanation:
We are given:
x' = 14 years
μ = 12.3 years
s = 0.8 years
Thus, let's use the formula for the Z-score value which is;
z = (x' - μ)/s
Thus;
z = (14 - 12.3)/0.8
z = 2.125
From the z-distribution table attached, the p-value is ;
P(x' > 2.125) = 1 - 0.983 = 0.017 = 1.7%
Thus,probability of lasting longer = 1.7%
The amount of gas consumed by first and second car were 20 gallons and 15 gallons respectively.
<em><u>Explanation</u></em>
Suppose,
gallons of gas were consumed by the first car.
As the total gas consumption in one week is 35 gallons, so the amount of gas consumed by second car will be:
gallons.
The first car has a fuel efficiency of 35 miles per gallon of gas and the second has a fuel efficiency of 15 miles per gallon of gas.
So, the <u>distance traveled by the first car</u> in
gallons of gas
miles and the <u>distance traveled by the second car</u> in
gallons of gas
miles.
Given that, the two cars went a <u>combined total of 925 miles</u>. So, the equation will be.....

So, the amount of gas consumed by the first car is 20 gallons and the amount of gas consumed by the second car is: (35 - 20) = 15 gallons.
Answer: 1.303639
Step-by-step explanation:
The t-score for a level of confidence
is given by :_
, where df is the degree of freedom and
is the significance level.
Given : Level of significance : 
Then , significance level : 
Sample size : 
Then , the degree of freedom for t-distribution: 
Using the normal t-distribution table, we have

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639