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Fofino [41]
3 years ago
15

How many solutions does the system of equations have x-2y=6 and 3x-6y=18

Mathematics
2 answers:
Ymorist [56]3 years ago
8 0
Solve for x in X-2 (x=8)
Substitute X=8 into 3x - 6y =18 (24-6y=18)
Solve for y in 24-6y=18 (y=1)
X=8 and y=1
tigry1 [53]3 years ago
8 0
2y=6
3x-6y=18
Divide first equation by 2: y=3
Substitute y=3 into second equation: 3x-6(3)=18
Solve for x: 3x=36 x=12

Therefore there is only one solution: x=12 y=3


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A shopkeeper buys 10 television sets from a retailer at a rate of Euro 15,000 per set. He sells half of them at a profit of 25 %
Lapatulllka [165]

Answer: Profit of $3,750

Step-by-step explanation:

Sold half of the sets at a profit of 25%;

= (15,000 * 5 sets) * 25%

= $‭18,750‬

Sold the other half at a loss of 20%;

= (15,000 * 5) * 20%

= $15,000

Profit (loss) = 18,750 - 15,000

= $3,750

8 0
3 years ago
Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters
Mumz [18]

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.

Read more about extension lines

brainly.com/question/13362603

#SPJ1

8 0
2 years ago
1. Emily wants to buy turquoise stones on her trip to New Mexico to give to at least 4 of her friends. The
Akimi4 [234]

Answer: she could get any of them the smalls ones or the large ones because 4x4=16 and 4x6=24 so she could pick the small or the large ones.

8 0
3 years ago
The standard golf ball has a mass of 45g.How many golf balls are there in a bag weighing 1.8 kg?
mars1129 [50]
Bag weight 1.8 kg
Golf ball weight: 45g

1 kg = 1000g
Total bag weight in grams = 1.8 x 1000 = 1800g
total bag weight / golf ball weight = total balls in bag
1800 / 45 = 40
Total balls in bag = 40.
6 0
3 years ago
Read 2 more answers
Simplify the logarithm log2(32)
Rom4ik [11]

let log2 32 = x

then 2^x = 32

but we know 2^5 = 32

therefore, x = 5

6 0
3 years ago
Read 2 more answers
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