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3 years ago

Find the derivative of y = ln(cx).

Mathematics

1 answer:

kenny6666 [7]3 years ago

8 0

Answer:

y'=1/x

Step-by-step explanation:

$d/dx[ ln(u) ]=\frac{1}{u} *du\\y'=\frac{1}{cx} *c\\y'=\frac{1}{x}$

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I need help.What do I do?

iris [78.8K]

4^2+4×11
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60 is the answer to the question

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3 years ago

Read 2 more answers

the figure is made up of a semicircle and a triangle. Find the area of the figure. Use pi equals 3.14

valentinak56 [21]

Answer:

The area of the figure is 63.25 square feet

Step-by-step explanation:

we know that

The area of the figure is equal to the area of semicircle plus the area of triangle

step 1

Find the area of semicircle

The area of semicircle is

$A_1=\frac{1}{2}\pi r^{2}$

we have

$r=10/2=5\ ft$ ----> the radius is half the diameter

$\pi =3.14$

substitute

$A_1=\frac{1}{2}(3.14)5^{2}$

$A_1=39.25\ ft^2$

step 2

Find the area of the triangle

The area of triangle is equal to

$A_2=\frac{1}{2}bh$

we have

$b=8\ ft\\h=6\ ft$

substitute

$A_2=\frac{1}{2}(8)(6)$

$A_2=24\ ft^2$

step 3

The area of the figure is equal to

$A=A_1+A_2$

substitute

$A=39.25+24=63.25\ ft^2$

5 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.

Read more about extension lines

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1 year ago

A van drove 241. 92 miles on 10. 8 gallons of gas. How far could the van drive on a full tank of 16. 2 gallons of gas? Drag and

mrs_skeptik [129]

You can use the given data to find out how many miles does the van travels for 1 gallon of gas. This will help you get the distance van can travel for full tank (16.2 gallons) of gas.

The considered van with full tank(16.2 gallons) can travel 362.88 miles

<h3>How to find the distance the van can travel per gallon of gas?</h3>

Since the considered van traveled 241.92 miles in 10.8 gallons, then we can take equal in dimension of input and output relation.

$10.8 \: \rm gallons = 241.92 \: miles\\\\\text{Dividing both quantities by 10.8 to get miles traveled per gallon}\\\\\dfrac{10.8}{10.8} \: gallon = \dfrac{241.92}{10.8} \: miles\\\\ 1 \: gallon = 22.4 \: miles$

Thus, the considered van travels 22.4 miles per gallon of gas.

Now, since full tank consists of 16.2 gallons, thus, multiplying by 16.2 on both the sides, we get

$\rm 1 \: gallon = 22.4 \: miles\\\\16.2 \: gallons = 16.2 \times 22.4 \: miles = 362.88 \: miles$

Thus,

The considered van with full tank(16.2 gallons) can travel 362.88 miles

Learn more about division here:

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2 years ago

You added 3 3/4 cups of flour, 1 1/4 cups of oil, and 2 2/4 cups of water to a bowl for a recipe. How many cups did you combine

Archy [21]

3 3/4 + 1 1/4 + 2 2/4

3/4 + 1/4 + 2/4 = 1 1/2

3 + 1 + 2 + 1 1/2 = 7 1/2 cups total

4 0

3 years ago