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3 years ago

13

In a downtown parking lot there are 5 times as many cars there as trucks. There are 12 trucks in the parking lot. How many cars

are in the parking lot

2 answers:

Alecsey [184]3 years ago

8 0

There are 60 cars in the parking lot

5 times 12 trucks=60

Hope that helped :)

torisob [31]3 years ago

7 0

12 * 5 = 60 Cars & 12 Trucks

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What is the simplified form of the expression? 9. (m^7)^2 a) 2m^14 b) m^49 c) m^9 d) m^14

kupik [55]

(m^7)^2 = m^(7 x 2) = m^14

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3 years ago

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

s344n2d4d5 [400]

The various answers to the question are:

To answer 90% of calls instantly, the organization needs four extension lines.
The average number of extension lines that will be busy is Four
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$$

Generally, the equation for is mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

$P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

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3 0

1 year ago

7. Susie bought some reams of paper for 5 each and a $200 printer. He spent a minimal of $450. write and solve an equation to fi

Dovator [93]

Answer:

$450-200÷5=the number of reams of paper Susie purchased.

Step-by-step explanation:

$450-$200=$250

$250÷5= 50

Therefore, Susie purchased 50 reams of paper

3 0

2 years ago

Allie deposited $80 in a savings account earning 10% interest, compounded annually. To the nearest cent, how much will she have

77julia77 [94]

Answer: $96.8

Step-by-step explanation:

Formular for Compounding annually :

A =P ( 1 + r/n) ^nt

Where p is the Principal given as $80

r is the rate given as 10% = 10/100 = 0.1

n is the number of times the principal was compounded, given as 1

t is the time, given as 2 years:

Slot the values into the formula:

A= $80 ( 1 + 0.1/1) ^ (1 × 2)

A= $80 ( 1 + 0.1) ^2

A= $80 ( 1.1) ^2

A= $80 (1.21)

A= $80 x 1.21

A= $96.8

96.8 is already rounded to the nearest cent.

5 0

3 years ago

A stained glass window is going to be installed in a semi-circular opening, which is above a 30 inch wide door. If the stained g

xenn [34]

The answer is <span>$494.55</span>

Let's first imagine a circle and calculate its area and then reduce it in half for the area of a semi-circle. Since this opening is above <span>a 30-inch wide door, the circle will have a diameter of 30 inches.
The area of the circle (A) is:
A = </span>π · r²
where:
π = 3.14
r - radius: r = diameter ÷ 2 = 30 ÷ 2 = 15 inches

So, the area of the circle is:
A = π · r² = 3.14 · 15² = 706.5 inches²

The area of the semicircle is half of the area of the circle:
A1 = A ÷ 2 = 706.5 ÷ 2 = 353.25 inches²

Since the stained glass window costs $1.40 <span>per square inch, for 353.25 square inches it will cost $494.55:
353.25 square inches * 1.40 $/square inch = $494.55</span>

4 0

3 years ago