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Serhud [2]
3 years ago
12

Later, the teaching assistant in Bob's chemistry course gives him some advice. "Based on past experience," the teaching assistan

t says, "working on 70 problems raises a student's exam score by about the same amount as reading the textbook for 1 hour." For simplicity, assume students always cover the same number of pages during each hour they spend reading.
Given this information, in order to use his 4 hours of study time to get the best exam score possible, how many hours should he have spent working on problems, and how many should he have spent reading?

1 hour working on problems, 3 hours reading

2 hours working on problems, 2 hours reading

3 hours working on problems, 1 hour reading

4 hours working on problems, 0 hours reading
Business
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

1 hour working on problems, 3 hours reading

Explanation:

the question is not complete:

Bob is a hard-working college freshman. One Tuesday, he decides to work nonstop until he has answered 200 practice problems for his chemistry course. He starts work at 8:00 AM and uses a table to keep track of his progress throughout the day. He notices that as he gets tired, it takes him longer to solve each problem.

Time                   Total Problems Answered       Marginal gain

8:00 AM                         0    

9:00 AM                        80                                        80

10:00 AM                     140                                        60

11:00 AM                      180                                        40

Noon                          200                                        20

Since Bob's is able to answer more than 70 questions per hours only during one hour (from 8 to 9 AM), he can benefit more from reading the next 3 hours. Reading would be equivalent to answering 210 questions, while Bob was only able to answer 120 more questions in the following 3 hours.

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The Western and Pacific Railroad has two divisions, the Western Division and the Pacific Division. The company recently invested
erik [133]

Answer:

$3,286,722

Explanation:

Allocate the improvement cost on the basis of <em>miles traveled</em> by the respective divisions.

<u>Western Division </u>

improvement cost = (890,000 / (890,000+1,520,000)) × $8,900,000

                              = 890,000 / 2,410,000 × $8,900,000

                              = $3,286,722

7 0
4 years ago
A quality-control technician has been monitoring the output of a milling machine. Each day, the technician selects a random samp
aleksandrvk [35]

Answer: 1.292 millimeters

Explanation:

Given the following ;

Number of Random sample (n) = 20

Average diameter(mean) = 1.327 millimeters

Standard deviation(std) = 0.0525 millimeters

To calculate the Lower Control Limit (LCL) :

Recall:

LCL = mean - [3(std÷sqrt(n))]

LCL = 1.327 - [3(0.0525÷sqrt(20))]

LCL = 1.327 - [3(0.0525 ÷ 4.472135955)]

LCL = 1.327 - [3 × 0.0117393568818]

LCL = 1.327 - 0.0352180706456216

LCL = 1.2917819293543784

LCL = 1.292 millimeters

6 0
4 years ago
Sam has been saving his money to purchase a new computer. He believes he has $781.70 in his checking account, but when he uses h
Lelu [443]
  1. Yeah, Sam hadn't saved enough money  
  2. During his estimates, Sam made some mistakes, but he had less capital on his record than  
  3. Sam made a few mistakes during his reports because he had less money than he expected on his account

<u>Explanation: </u>

A financial transaction is an arrangement or a contract to swap goods for compensation between a purchaser and a seller. There is a transition in the financial situation of two or more companies or persons.

A cashless company describes an economy in which financial transactions consist entirely of electronic data (generally an electronic portrayal of money) among transacting parties rather than money in a form of personal banknotes or gold and silver.

The trade between the Organization and another individual is an official contract. A great example of an extrinsic transaction is the purchase of goods from either a third party seller. An overall journal entry records each cash payment in the billing system.

8 0
3 years ago
Horton invests personally owned equipment, which originally cost $110,000 and has accumulated depreciation of $30,000 in the Hor
Zarrin [17]

Answer:

Dr Equipment $60,000

Cr. Horton, capital $60,000

Explanation:

Based on the information given we were told that Both of the partners agree that the fair value of the equipment was the amount of $60,000 which means that The appropiate journal entry made by the partnership to record Horton's investment should be:

Dr Equipment $60,000

Cr. Horton, capital $60,000

6 0
3 years ago
Larry Ellison starts a company that manufactures high-end custom leather bags. He hires two employees. Each employee only begins
HACTEHA [7]

Answer:

12.55 days

Explanation:

<em><u>Provided information </u></em>

Number of employees 2

Average production time=1.8 days

Standard deviation=2.7 days

Inter-arrival time= 1 day

Coefficient of variation= 1 day

Standard deviation of inter-arrival time= 1 day

The coefficient of variations

<u>Inter-arrival coefficient of variation </u>

C_{vi}=\frac {\sigma}{T} where \sigma is standard deviation of inter-arrival time, T is inter-arrival time and C_v is coefficient of variation of inter-arrival time

C_{vi}=\frac {1 day}{1 day}=1

<u>Production time coefficient of variation </u>

C_{vp}=\frac {2.7}{1.8}=1.5

<u><em>Total utilization time </em></u>

U=\frac {T}{n*T_i} where T is the time of production, n is number of employees, U is utilization, T_i is inter-arrival time

U=\frac {1.8}{2*1}=0.9

Therefore, utilization time by 2 employees is 0.9

<u>Expected average waiting time </u>

T_e=(\frac {T}{n*T_i})*0.5(C_{vi}^{2}+C_{vp}^{2})*(\frac{U^{\sqrt{2(n+1)}-1}}{1-U})

Where T_e is expected average waiting time and the other symbols as already defined

Substituting 1.5 for C_{vp}, 1 for C_{vi}, 0.9 for U, 2 for n, 1 for T_iand 1.8 for T

T_e=(\frac {1.8}{2*1})*0.5(1^{2}+1.5^{2})*(\frac{0.9^{\sqrt{2(2+1)}-1}}{1-0.9})

T_e=0.9*1.625*8.583709=12.55367 days  and rounding off to 2 decimal places we obtain 12.55 days

Therefore, expected duration between order received and beginning of production is approximately 12.55 days

4 0
3 years ago
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