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3 years ago

Which equation correctly shows how to determine the distance between the points (9, –2) and (6, 3) on a coordinate grid?

Mathematics

2 answers:

aleksley [76]3 years ago

8 0

Answer:

Third option :)

Step-by-step explanation:

nalin [4]3 years ago

4 0

Answer:

The formula for calculate the distance (d) between two points is: Given the points (9,-12) and (6,3), you can identify: Substitute these coordinates into the formula to find the equation that correctly shows how to determine the distance between the given points (9,-12) and (6,3)

Step-by-step explanation:

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Tia cut a 4 meter 8 centimeter wire into 10 equal pieces. Marta cut a 540 centimeter wire into 9 equal pieces. how much longer i

Molodets [167]

One meter is 100 meters so 4 meter 8 centimeter is 400 centimeter + 8 centemeter=408. if we cut that into 10 equal pieces, we get 40.8. Marta has 540 centimeters to cut into 9 pieces. 540/9=60. 60-40.8=19.2 marta's wire is 19.2 centimeters longet than tia's

7 0

3 years ago

Read 2 more answers

HELP PLS use the original price and markup to find the retail price. original price $68 markup 20%

insens350 [35]

Answer: $81.60

Step-by-step explanation:

When we have an original price OP, and a markup of X% (this is how much increases the price with respect to the original price)

The retail price will be:

RP = OP + (X%/100%)*OP

RP = OP*(1 + X%/100%)

In this case, we have:

Original price = OP = $68

markup = X% = 20%

Replacing these in the above equation, we get:

RP= $68*(1 + 20%/100%) = $68*(1 + 0.20)

RP = $68*(1.20) = $81.60

Then the retail price of this particular item is $81.60

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%20%7Bw%7D%5E%7B6%7D%20%20%7Bv%7D%5E%7B5%7D%20%7D%7B15%20%7Bw%7D%5E%7B2%7D%20v%7

ehidna [41]

Breaking down:
$\frac{3}{15} = \frac{1}{5}$
${w}^{6} \div {w}^{2} = {w}^{4}$
${v}^{5} \div v = {v}^{4}$

Therefore,
$\frac{3 {w}^{6} {v}^{5} }{15 {w}^{2}v } \\ = \frac{{w}^{4} {v}^{4} }{5}$

Hope this helps. - M

5 0

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 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.

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

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.