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3 years ago

If 4a - 3 = 9, what is the value of 4a + 1?! A: 13 B: 3 C:49 D: 5

Mathematics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:

A: 13

Step-by-step explanation:

4a-3 = 9

add 3 to both sides

4a-3 = 9

+3 +3

the -3 and +3 will cancel each other out

and from adding the 3 to the 9 will turn the 9 into 12!

Now your equation will look like this

4a = 12

Next divide both sides by 4. the 4 will cancel out the other 4.\

now you will have your answer

a = 3

so now substitute the A with 3 in

4a+1

so it will be

4(3)+1

4 times 3 is equal to 12

then 12+1 = 13

I hope this helps! :)

Oliga [24]3 years ago

6 0

Answer:

a=3

Step-by-step explanation:

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The perimeter is 20 because you just count how many squares are touching the sides

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3 years ago

Write the equation of a line that passes through (3.5,0) and is perpendicular to 8y+4x=64.

pantera1 [17]

Answer:

y = 2x + 3.5

Step-by-step explanation:

Step 1: find the slope

8y + 4x = 64

8y = 64 - 4x

Make y the subject of the formula

y = (64 - 4x)/8

y = ( -4x + 64)/8

Separate to get slope

y = -4x/8 + 64/8

y = -x/2 + 8

Slope is the coefficient of x

m = -1/2

Note: if two lines are perpendicular to the other , it is negative reciprocal to each other

m = 2

Using the point slope form equation

y - y1 = m(x - x1)

y - y1 = 2(x - x1)

Substitute the point

( 3.5 , 0)

x1 = 3.5

y1 = 0

y - 3.5 = 2( x - 0)

open the bracket

y - 3.5 = 2x - 0

y = 2x - 0 + 3.5

y = 2x + 3.5

The equation of the line is

y = 2x + 3.5

7 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 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.