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3 years ago

6: the cost of a protein bar increased from $2.50 to $2.80. the percent increase in the $2.80 rate was how much?

1 answer:

5 0

<span>The original cost of protein bar = $2.50
The increased cost of protein bar = $2.80
Therefore, the increase in price of the protein bar = $2.80-$2.50 =$0.30 percentage increase = (Difference in price / Original price) x 100
=(0.30/2.50)x100
=(3/25)x100
=3 x 4
=12%</span>

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Answer: $428,000

Explanation:

Given that,

Accounts payable = $62,000

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Total assets = Accounts receivable + Cash + Inventory + Land

= 100,000 + 30,000 + 138,000 + 160,000

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Answer:

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Helen [10]

Question Completion:

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Check figures:

· Net income is $3,800.

· Total assets are $37,000.

· Total liabilities are $12,500.

Answer:

Elite Service Co.

a) Income Statement for the month of June 30, 2022:

Service revenue $7,500

Supplies expense $1,000

Maintenance and repairs expense 600

Advertising expense 400

Utilities expense 300

Salaries and wages expense 1,400 3,700

Net Income $3,800

b) Retained Earnings Statement for the month of June 30, 2022:

Net Income $3,800

Dividends paid 1,400

Retained Earnings $2,400

c) Balance Sheet as of June 30, 2022:

Cash $ 4,600

Accounts receivable 4,000

Supplies 2,400

Equipment 26,000

Total assets $37,000

Notes payable $12,000

Accounts payable 500

Total liabilities $12,500

Common stock 22,100

Retained Earnings $2,400

Total liabilities + Equity $37,000

Explanation:

Data and Calculations:

Cash $ 4,600

Accounts receivable 4,000

Supplies 2,400

Equipment 26,000

Total assets $37,000

Notes payable $12,000

Accounts payable 500

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2 years ago

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3 years ago

Agan Interior Design provides home and office decorating assistance to its customers. In normal operation, an average of 2.5 cus

Phantasy [73]

Answer:

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$\lambda=2.5 \,customers/hour\\\\\mu=6\,customers/hour$

B) The goal is not met, as the average time waiting for service is 5.56 minutes.

C) The new mean service rate is 7.5 customers/hour.

In this case, the average time waiting for service is 4 minutes, so the goal is met.

Explanation:

A) This situation can be modeled as a single-server single-phase model (M/M/1).

The mean arrival rate is 2.5 customers per hour.

$\lambda=2.5 \,customer/h$

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$\mu=\frac{60\, min/h}{10 \,min/customer}=6\, customer/h$

B) The average waiting time for a customer can be expressed as:

$W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda} =\frac{2.5}{6}\frac{1}{6-2.5} =0.417*0.222=0.093\,hours\\\\W_q=0.093\,hours*(60min/h)=5.56 \,min$

The average waiting time is 5.56 minutes, so it is more than the goal of 5 minutes.

C) If the average time spent per customer to 8 minutes, the mean service rate becomes

$\mu=\frac{60\, min/h}{8 \,min/customer}=7.5\, customer/h$

An the average waiting time for the service now becomes:

$W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda} =\frac{2.5}{7.5}\frac{1}{7.5-2.5} =0.333*0.2=0.067\,hours\\\\W_q=0.067\,hours*(60min/h)=4 \,min$

The average time is now 4 minutes, so the goal is achieved.

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3 years ago