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Nadya [2.5K]
3 years ago
13

What is 3/4 + 1/8 sorry i didnt know this im really needing help with this anwser

Mathematics
2 answers:
Luden [163]3 years ago
8 0

Answer:

<h2>You answer is ⅞.</h2>

Step-by-step explanation:

\frac{3}{4}  +  \frac{1}{8}

To solve this question we need to make the denominators same.

We can make it by finding the L.C.M.

We know,

The L.C.M of 4 and 8 is 8.

So therefore the fractions will become -

\frac{3}{4}  =  \frac{3 \times 2}{4 \times 2}  =  \frac{6}{8}

And

\frac{1}{8}  =  \frac{1 \times 1}{8 \times 1}  =  \frac{1}{8}

Now we can add the numbers

\frac{6}{8}  +  \frac{1}{8}  =  \frac{6 + 1}{8}  =  \frac{7}{8}

So the answer is ⅞.

Hope it helps you!!

#IndianMurgaツ

saul85 [17]3 years ago
3 0

Answer:

7/8

Step-by-step explanation:

hope this helps :)

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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

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8 0
2 years ago
Rasheed is buying wings and quesadillas for a party. A package of wings costs $8. A package of quesadillas costs $10. He must sp
galina1969 [7]

Answer:

10 is maximum of packages of quesadillas.

Step-by-step explanation:

x= number of wings

y= number of quesadillas

x+y=20

8x+10y=$160

if x=7

then 7+y=20

20-7=14=y

14*10=140

8*7=56

Rasheed can't buy 20 packages, since it'll go over his budget.

So 160-56=104

And 104/10= 10 minimum

10*10=100

100+56=160

So he can buy 10 packages of quesadillas.

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vovangra [49]

Answer:

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Step-by-step explanation:

The given statement implies that the slope of line formed by (5, r) is (3, -2) is -2, as they lie on the same line.

Slope(using these pts) = (y2 - y1)/(x2 - x1)

=> - 2 = (-2 - r)/(3 - 5)

=> -2 = (-2 - r)/(-2)

=> (-2)(-2) = (-2 - r)

=> 4 = -2 - r

=> r = - 2 - 4

=> r = - 6

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The diameter of a Frisbee is 12 in. What is the area of the Frisbee?
Alina [70]

we know that

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3 years ago
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Over [174]

Answer:

84 cups of water will leak out after 2 weeks

Step-by-step explanation:

* Lets how many how to change oz to cup

- One cup contains 8 oz

- We can use the ratio to solve this problem

- A hose is leaking water at a rate of 2 oz/h

- We need to find how many cups of water will leak out after 2 weeks

∵ The rate of the water leak out is 2 oz/hr

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∴ 1 oz = \frac{1}{8} cup

∴ 2 oz = 2 × \frac{1}{8} = \frac{2}{8} = \frac{1}{4} cup

∴ The rate of the water leak out = \frac{1}{4} cup/h

∵ 1 day = 24 hours

∵ 1 week = 7 days

∴ 1 week = 7 × 24 = 168 hours

∵ The number of hours in 1 week is 168

∴ The number of hours in 2 weeks = 168 × 2 = 336 hours

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⇒      \frac{1}{4}         :   1

⇒     x         :    336

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∴ x = 336 × \frac{1}{4}

∴ x = 84

- x represents the number of cups of water leak out

<em>84 cups of water will leak out after 2 weeks</em>

5 0
4 years ago
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