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3 years ago

10

Sharon needs 64 credits to graduate from her community college. So far she has earned 24 credits. What percent of the required c

redits does she have?

2 answers:

Alina [70]3 years ago

6 0

She has 37.5 % of her answers because 64/124 is 37.5

Roman55 [17]3 years ago

4 0

37.5% of her credits. 24 out of 64, divide and you get your answer. multiply by 100 to get the percentage.

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Write the equation in slope-intercept form.<br><br> y=12+(x-1)(-4)

Alex Ar [27]

To get slope-intercept form we have to multiply x-1 by -4 and simplify it then
y=12+(x-1)(-4)
y=12+x*(-4)-1*(-4)
y=12-4x+4
y=-4x+16 - its the result

3 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

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8 0

1 year ago

Square root 230in simplified in redical form.

Mariulka [41]

Ok so you find squares
example
we know ataht √9=3 becuase 9=3*3 so √9=√(3 times 3)=√(3^2) =3

so we factor 230
230=2 times 5 times 23
we do not have any doubles so this is simplest form

6 0

3 years ago

Use the order pairs to write a function rule. Give the tule in slope-intercept form.

topjm [15]

to get the equation of any straight line, we simply need two points off of it, so hmmm let's use say (-1 , -1.25) and (8 , -3.5)

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1.25})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-3.5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3.5}-\stackrel{y1}{(-1.25)}}}{\underset{run} {\underset{x_2}{8}-\underset{x_1}{(-1)}}}\implies \cfrac{-3.5+1.25}{8+1}\implies \cfrac{-2.25}{9}\implies -\cfrac{~~ \frac{225}{100}~~}{\frac{9}{1}} \\\\\\ -\cfrac{225}{100}\cdot \cfrac{1}{9}\implies -\cfrac{9}{4}\cdot \cfrac{1}{9}\implies -\cfrac{1}{4}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1.25)}=\stackrel{m}{-\cfrac{1}{4}}(x-\stackrel{x_1}{(-1)}) \\\\\\ y+1.25=-\cfrac{1}{4}(x+1)\implies y+\cfrac{5}{4}=-\cfrac{1}{4}x-\cfrac{1}{4} \\\\\\ y=-\cfrac{1}{4}x-\cfrac{1}{4}-\cfrac{5}{4}\implies y=-\cfrac{1}{4}x-\cfrac{6}{4}\implies y=-\cfrac{1}{4}x-\cfrac{3}{2}$

3 0

2 years ago

How can a Line Plot help you find an average with data given in Fractions?

umka2103 [35]

Answer:

I did not understand its confusing

7 0

2 years ago

Read 2 more answers