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3 years ago

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Which is equivalent to

ula1" title="\sqrt[4]{9} ^{\frac{1}{2}x }" alt="\sqrt[4]{9} ^{\frac{1}{2}x }" align="absmiddle" class="latex-formula">
A: $9^{2x}$
B: $9^{\frac{1}{8}x }$
C: $\sqrt{9}^{x}$
D: $\sqrt[6]{9} ^{x}$

ANSWER IS B: : $9^{\frac{1}{8}x }$

1 answer:

Digiron [165]3 years ago

5 0

For this case we have that by definition of properties of powers and roots, it is fulfilled that:

$\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}$

So:

$\sqrt [4] {9 ^ {\frac {1} {2} x}} = 9 ^ {\frac {\frac {1} {2}} {4} x} = 9 ^ {\frac {1} {8} x}$

So, we have to:

$\sqrt [4] {9 ^ {\frac {1} {2} x}} = 9 ^ {\frac {1} {8} x}$

Answer:

$9 ^ {\frac {1} {8} x}$

Option B

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Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

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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.

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2 years ago

if a shoelace drawer is 12.49 inches and a paperclip drawer is 4.579 inches about how many inches wider is the shoelace drawer t

sammy [17]

The shoelace drawer is 7.911 inches wider than the paperclip drawer.

<h3>Measurement difference between two drawers</h3>

The height of shoelace drawer is = 12.49 inches

The height of paperclip drawer is = 4.579 inches

Therefore the measurement difference between the two drawer is = 12.49 inches - 4.579 inches

= 7.911 inches

Therefore, the shoelace drawer is 7.911 inches wider than the paperclip drawer.

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2 years ago

Determine the domain of the function (fog) (x) where f(x) = 3x-1/x-4 and g(x)=x+1/x

Mrac [35]

Answer:

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

Step-by-step explanation:

The given functions are $f(x)=\frac{3x-1}{x-4}$ and $g(x)=\frac{x+1}{x}$ .

We now composed the two functions to find:

$(f\circ g)(x)=f(g(x))$

$\implies (f\circ g)(x)=f(\frac{x+1}{x})$

$\implies (f\circ g)(x)=\frac{3(\frac{x+1}{x})+1}{\frac{x+1}{x}-4}$

$\implies (f\circ g)(x)=\frac{4x+3}{1-3x}$

This function is defined if the denominator is not zero.

$1-3x\ne0$

$x\ne\frac{1}{3}$

We write this in interval notation as:

$(-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty)$

We need to be cautious here as x=0 is not in the domain of g(x).

Therefore the domain of $(f\circ g)(x)$ is

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

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3 years ago

The probability of picking two black marbles from a box at random without replacement is 10/91 .

sasho [114]

The probability of the 2 events are multiplied to get the result 10/91

so we have the relation 5/14 * x = 10/91

where x is the probability of drawing the second black

so x = 10/91 * 14/5 = 4/13 answer

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3 years ago

Read 2 more answers

Suppose James has a credit card with a balance of 4289 . Each month the credit card company charges 5% interest. If James pays o

topjm [15]

Answer:

<h2>4076.56</h2>

Step-by-step explanation:

First we need to calculate the James monthly charges on his balance of 4289.

Using the simple interest formula;

Simple Interest = Principal * Rate * Time/100

Principal = 4289

Rate = 5%

Time = 1 month = 1/12 year

Simple interest = 4289*5*1/12*100

Simple interest = 21,445/1200

Simple interest = 17.87

<u>If monthly charge is 17.87, yearly charge will be 12 * 17.87 = </u><u>214.44</u>

The balance on his credit card one year from now = Principal - Interest

= 4289 - 214.44

= 4076.56

The balance on his credit card one year from now will be 4076.56

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4 years ago