Barker and Associates specialize in corporate law. They charge $125 an hour for researching a case, $80 an hour for consultations, and $250 an hour for writing a brief. Last week one of the attorneys spent 8 hours consulting with her client, 12 hours researching the case, and 15 hours writing the brief. What was the weighted mean hourly charge for her legal services
Answers
Answer:
Weighted mean hourly charge = $168.28 (Approx)
Given:
Charge for research = $125 per hour
Charge for consultations = $80 per hour
Charge for writing a brief = $250 per hour
Research work = 12 hour
Consultations = 8 hour
Writing = 15 hour
Computation:
Total hours = 12 + 8 + 15 = 35 hours
Total charge for researching = $125 × 12 = $1,500
Total charge for consulting = $80 × 8 = $640
Total charge for writing = $250 × 15 = $3,750
Total charge = $5,890
Weighted mean hourly charge = 5890 / 35
Weighted mean hourly charge = $168.28 (Approx)
Related Questions
Write and evaluate the expression. Then, complete the statements. nine more than the product of seven and a number decreased by three The expression to model the situation is 1. 3 - 9(7n) 2. 9 + 7/n - 3 3. 9 + (7 + n) - 3 4. 9 + 7n - 3 . The value of the expression when n = 8 is 1. 6 7/8 2. 71 3. 62 4. 501
Mr. Bruce has some clay with a mass of 24 kilograms. If he puts the same amount of clay into 6 bags, how many grams of clay are in each bag? I NEED THE EXPLANATION
HELP ME OUT PLEASE!!! In triangle ABC, the length of side AB is 15 inches and the length of side BC is 21 inches. Which of the following could be the length of side AC?A. 41 inchesB. 4 inchesC. 38 inchesD. 30 inches
A painting, square in shape, is placed in a wooden frame with width of 10% of the length of the side of the paining. The painting was enlarged by 10%. By what percent is the new frame bigger than the original frame if the width of the frame remains the same?
Answers
The answer is 70.......
Answer:
28% of 250 is 70
Step-by-step explanation:
If you turn 28% into a decimal it will turn into 0.28 so then multiply that by 250 and you'll get 70
Answers
The perimeter and area of the composite shape is:
- B. Perimeter = 19.42 inches; Area = 26.13 square inches
Recall:
Area of a circle = πr²
Perimeter of circle = 2πr
Area of triangle = 1/2(bh)
The composite shape given is composed of a triangle and a semicircle.
Perimeter of the composite shape = Perimeter of semicircle + the length of the two sides of the triangle
Perimeter = 1/2(2 × 3.14 × 3) + 2(5) = 19.42 inches
Area of the composite shape = area of semicircle + area of triangle
Area = 1/2(3.14 × 3²) + 1/2(6 × 4)
Are = 14.13 + 12
Area of the composite shape = 26.13 square inches.
Therefore, the perimeter and area of the composite shape is:
- B. Perimeter = 19.42 inches; Area = 26.13 square inches
Learn more about area and perimeter of composite shapes on:
Answer:
Step-by-step explanation:
The composite shape consists of a semi circle and a triangle. The formula for determining the perimeter of a semicircle is expressed as
Perimeter = 1/2 × 2πr = πr
Since radius, r = 3, then
Perimeter of semi circle = 3 × 3.14 = 9.42 inches
Perimeter of composite shape = 9.42 + 5 + 5 = 19.42 inches
Area of semi circle = 1/2 × πr²
Area of semicircle = 1/2 × 3.14 × 3² = 14.13 inches²
Area of triangle = 1/2 × base × height
Area of triangle = 1/2 × 6 × 4 = 12 inches²
Area of composite shape = 14.13 + 12 = 26.13 inches²
Answers
P(A) = 2/5 since there are 2 black out of 2+3 = 5 total
After you make a selection, we have the event
B = event that you select another black sweater assuming event A has happened already
P(B) = 1/4 because there's 1 black sweater left out of 5-1 = 4 left over
Multiply the probabilities
P(A and B) = P(A)*P(B)
P(A and B) = (2/5)*(1/4)
P(A and B) = 2/20
P(A and B) = 1/10
The answer as a fraction is 1/10
In decimal form, it is 0.1
As a percent, the answer is 10%
Answers
For a vending machine having Service time is 20 seconds per cup and customers arrive at a mean rate of 64 per hour, then average number of customers waiting in a line is 0.10
Number of customer in a queue means those who are waiting for a server.
Given the following information:
Mean arrival rate of customer, μ=64 customers per hour
Service time is 20 seconds per cup that is 1 customer per 20 seconds
λ=180 customers per hour
Average number of customers waiting in a line,
On substituting the values,
Thus, average number of customers waiting in a line is 0.10
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Complete question:
A vending machine dispenses hot chocolate or coffee. Service time is 20 seconds per cup and is constant. Customers arrive at a mean rate of 64 per hour, and this rate is Poisson-distributed. Determine the average number of customers waiting in line.
Final answer:
This problem engages queueing theory in mathematics, specifically it involves a vending machine with constant service time and Poisson-distributed customer arrival rate. The system is analyzed to be stable as the service rate surpasses the arrival rate.
Explanation:
This problem is a classic case of queueing theory in mathematics, particularly relevant in Probability and Statistics. Our case involves a vending machine that has a constant service time of 20 seconds per cup of hot chocolate or coffee. The mean customer arrival rate is presented as 64 per hour, described as being Poisson-distributed.
To start, consider the service rate. With the service time being a constant 20 seconds per cup, this translates to 3 cups being served per minute or 180 cups per hour. This value becomes our service rate µ. For the arrival rate or lambda (λ), the rate was given as 64 customers per hour.
In this particular queuing system, the service rate is higher than the arrival rate. This means that the system is stable, and queues are not expected to be overly long because customers are being served at a faster rate than they are arriving.
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B) cos(C)
C) sin(B)
D) sin(C)
Answers
cosA=AC/AB
sinB=AC/AB
hence cosA=sinB
Answer:
C).
Step-by-step explanation:
Since angles A and B are complementary, their cofunctions are equal. So, cos(A) = sin(B).
Answers
Answer:
Step-by-step explanation:
There are lots of ways we can think about the typical number of cavities.
- What was the most common number of cavities?
- If we split the cavities evenly among all the patients, how many cavities would each patient have?
- What would be the balance point of the data?
- What is the middlemost number of cavities?
The most patients had 0cavities.
If we split the cavities evenly, each patient would have 2 or 3 cavities.
If we put our dot plot on a balance scale, it would balance when the pivot was between 2 and 3 cavities.
The scale would tip if, for example, we put the pivot at 5 cavities.
There are 8 patients with 2 cavities each. About half of the rest of the patients have fewer than 2 cavities and about half have more than 2 cavities.
Of the choices, it is reasonable to say that a patient typically had about 2 cavities.
,
-Written in
Final answer:
The 'typical' number of cavities one patient had can be determined by finding the mode (most common number) in the data set, which should be represented in the dot plot. To do this, one would count the number of dots at each value on the dot plot. The value with the most dots would be the 'typical' number of cavities.
Explanation:
The question is asking for a 'typical' number of cavities one patient had out of Dr. Vance's 63 patients. In statistics, a typical, or 'common', value can be shown by calculating the mode, which is the number that appears most frequently in a data set.
Unfortunately, the dot plot is missing from the information provided. However, to find the mode (or typical value) using a dot plot, you would typically count how many dots are at each value on the plot. The value with the most dots (indicating the most patients with that number of cavities) is the mode. This would be the 'typical' number of cavities a patient of Dr. Vance had last month.
Let's create a hypothetical scenario. If your dot plot looked like this:
- 0 cavities: 10 patients
- 1 cavity: 15 patients
- 2 cavities: 24 patients
- 3 cavities: 8 patients
- 4 cavities: 6 patients
The mode would be 2 cavities because 24 patients had this amount, more than any other amount. Therefore, the 'typical' number of cavities one patient had would be 2.
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