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Kazeer [188]
3 years ago
13

In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout time,

with a standard deviation of 3.1 minutes. A sample of 72 customers who used a cashier averaged 6.1 minutes with a standard deviation of 2.8 minutes.
(a) Can you conclude that the mean checkout time is less for people who use the self-service lane?
(b) Can you conclude that if everyone used the self-service lane, that the mean checkout time would decrease? Consider the number of items checked out when formulating your answer.
Mathematics
1 answer:
bulgar [2K]3 years ago
7 0

Answer:

Step-by-step explanation:

I sample :

Sample size n1 = 60

Sample average = x bar = 5.2 minutes

STd deviation sample = s1 = 3.1 minutes

II sample:

n2 = 72

y bar = 6.1

s2 = 2.8 minutes

Mean difference = -0.9

Std error for difference = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2} } \\=0.5187

t statistic = mean diff/se = -1.735

p value = 0.0848

Since p >0.05 we accept null hypothesis that there is no difference between the two averages

Conclusion:

a) We cannot conclude that the mean checkout time is less for people who use the self-service lane

b) NO, we cannot conclude that if everyone used the self-service lane, that the mean checkout time would decrease

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A carpool service has 2,000 daily riders. A one-way ticket costs $5.00. The service estimates that for each $1.00 increase to th
solmaris [256]

Answer:

Total number of riders that ride on carpool daily = 2000

Total Cost of one way ticket = $ 5.00

Total Amount earned if 2000 passengers rides daily on carpool = 2000 × 5

                                                                                                            = $10,000

If fare increases by $ 1.00

New fare = $5 + $1    

               = $6

Number of passengers riding on carpool = 2,000 - 100 = 1,900

If 1,900 passengers rides on carpool daily , total amount earned ,if cost of each ticket is $ 6 = 1900 × $6 = $11400

As we have to find the inequality which represents the values of x that would allow the carpool service to have revenue of at least $12,000.

For $ 1 increase in fare = (2,000 - 1 × 100) passengers

For $ x increase in fare, number of passengers = 2,000 - 100·x

                                                         = (2,000 - 100·x) passengers

New fare = 5 + x

New Fare × Final Number of passengers ≥ 12,000

(5+x)·(2,000 - 100 x) ≥ 12,000

5 (2,000 - 100 x) + x(2,000 - 100 x) ≥ 12,000

10,000 - 500 x + 2,000 x - 100 x² ≥ 12,000

100 - 5 x + 20 x - x² ≥ 120

- x² + 15 x +100 - 120 ≥ 0

-x² + 15 x -20 ≥ 0

x² - 15 x + 20 ≤ 0

⇒ x = 1.495

x ≥ $ 1.495, that is if we increase the fare by this amount or more than this the revenue will be at least 12,000 or more .

Also, f'(x) = 0 gives x = 7.5

⇒ The price of a one-way ticket that will maximize revenue is $7.50

7 0
2 years ago
My question got deleted so I'll post these into two separate
Agata [3.3K]

Answer:

see below

Step-by-step explanation:

1 million

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.031709792 years

Rounding to 3 decimal places

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50 years * 365 days/ 1 year * 24 hours/ 1 day * 3600 second / 1 hour    

1577000000 seconds    

46 inches * 1 ft/ 12 inches *  1 mile/5280 ft

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Answer:

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