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OlgaM077 [116]
3 years ago
10

£)

Business
1 answer:
aleksley [76]3 years ago
3 0

Answer:

The greatest number of mangoes which are to be taken out at a time from each basket so that both of them emptied simultaneously is the number of mangoes in each basket which is 120 mangoes for one basket and 168 mangoes for the other basket

Explanation:

Given that the number of mangoes in one basket = 120 mangoes

Also, the number of mangoes in another basket = 168 mangoes

The greatest number of mangoes, X and Y that are to taken out from each basket so that both of them will empty simultaneously is found as follows;

We note that the ratio of the number of mangoes in both baskets are;

120:168 = 5:7

Therefore, we have;

5 × Y = 120

Y = 20/5 = 24

Similarly, we have;

7 × X = 168

X = 168/7 = 24

We can take 5 mangoes from one basket and 7 mangoes from the other basket 24 times, for both mangoes to empty the same time

We can also take 5×12 = 60 mangoes twice from one basket and 7 × 12 = 84 mangoes twice from the other basket to empty the baskets

We can also take 120 mangoes one from one basket and 168 mangoes one from the other basket to empty the baskets.

Therefore, the greatest number of mangoes which are to be taken out at a time from each basket so that both of them emptied simultaneously is the number of mangoes in each basket which is 120 mangoes for one basket and 168 mangoes for the other basket.

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Use the data below to construct the advance/decline line for the stock market. Volume figures are in thousands of shares. (Do no
slamgirl [31]

Answer:

                Adv./Dec.               Cumulative

Monday               1                            1

Tuesday              2                            3

Wednesday         1                            4

Thursday             5                            9

Friday                   1                            10

Explanation:

Note: See the attached excel file for the construction of he advance/decline line for the stock market.

Download xlsx
3 0
3 years ago
The set of unofficial relationships among an organizations members is called a(n __________ structure.
mafiozo [28]
Unofficial relationships in organizations can be referred to as ad hoc structures.
4 0
3 years ago
An investment of 1 will double in 15.142 years at a force of interest, δ. An investment of 1 will increase to 3 in n years at th
vredina [299]

Answer:

24 years

Explanation:

Nper = 15.142

PMT = 0

PV = 1

FV = 2

Type = 0

Rate = Rate(Nper, pmt, -pv, fv, type)

Rate = Rate(15.142, 0, -1, 2, 0)

Rate = 4.68%

Rate = 4.68%

PMT = 0

PV = 1

FV = 3

Type = 0

NPER = NPER(rate, pmt, -pc, fv, type)

NPER = NPER(0.04684, 0, -1, 3, 0)

NPER = 24 years

6 0
3 years ago
Brad Edwards is earning $74,000 a year in a city located in the Midwest. He is interviewing for a position in a city with a cost
arsen [322]

Answer:

The correct answer is $81,400.

Explanation:

According to the scenario, the given data are as follows:

Current earning = $74,000

As Brad is searching for a city which is 10% higher than current city then to maintain same living he has to earn 10% more than he earns.

So, total earning needed = $74,000 + 10% of $74,000

  =  $74,000 + $7,400

  = $81400

Hence, the total earning brad needed is $81,400.

8 0
3 years ago
The Bijou Theater in Hermosa Beach, California, shows vintage movies. Customers arrive at the theater line at the rate of 100 pe
a_sh-v [17]

Answer:

<u>Task a: </u>

<u>What is the average customer time in the system?</u>

Answer: 30 seconds

<u>Task b:</u>

<u>What would be the effect on customer time in the system of having a second ticket taker doing nothing, but validations and card punching, thereby cutting the average service time to 20 seconds?</u>

Answer: 33.34 minutes

<u>Task c:</u>

<u>Would system waiting time be less than you found in b if a second window was opened with each service doing all three tasks?</u>

Answer: 16.67 minutes

Explanation:

<u>Task a:</u>

<u>What is the average customer time in the system?</u>

<u>Solution:</u>

Average customer time = 30 seconds per customer.

Total number of customers in an hour = 100.

Total time taken to attend customers = time per customer*total customers

= 30*100 = 3,000 seconds or 50 minutes.

So, 100 customers are attended in 50 minutes.

Average time = 50/100 = 0.5 minutes or 30 seconds

<u>Task b:</u>

<u>What would be the effect on customer time in the system of having a second ticket taker doing nothing, but validations and card punching, thereby cutting the average service time to 20 seconds?</u>

<u>Solution:</u>

If the average service time is reduced to 20 seconds, the customer time will reduce in the system.

For 100 customers, total time taken = 20*100 = 2,000 seconds or 33.34 minutes

Thus waiting time for the 100th customer has fallen from 50 minutes to 33.34 minutes.

<u>Task c:</u>

<u>Would system waiting time be less than you found in b if a second window was opened with each service doing all three tasks?</u>

<u>Solution:</u>

In this case, 100 customers will be divided into 1/3rd each i.e. 100/3 in each line.

So, waiting time for the 100/3th customer = 100/3*original time = 100/3*30 seconds = 1,000 seconds or 16.67 minutes.

Thus, within 16.67 minutes the line would be serviced.

Therefore, System waiting time will be less than as calculated in b.

3 0
4 years ago
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