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3 years ago

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What is the domain of the graph ?

1 answer:

marysya [2.9K]3 years ago

3 0

The domain of the graph is all real numbers because the value doesn’t have an end point.

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Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

Ostrovityanka [42]

Answer:

2.25

Step-by-step explanation:

The computation of the number c that satisfied is shown below:

Given that

$f(x) = \sqrt{x}$

Interval = (0,9)

According to the Rolle's mean value theorem,

If f(x) is continuous in {a,b) and it is distinct also

And, f(a) ≠ f(b) so its existance should be at least one value

i.e

$f^i(c) = \frac{f(b) - f(a)}{b -a }$

After this,

$f(x) = \sqrt{x} \\\\ f^i(x) = \frac{1}{2}x ^{\frac{1}{2} - 1} \\\\ = \frac{1}{2}x ^{\frac{-1}{2}$

$f^i(x) = \frac{1}{{2}\sqrt{x} } = f^i(c) = \frac{1}{{2}\sqrt{c} } \\\\\a = 0, f (a) = f(o) = \sqrt{0} = 0 \\\\\ b = 9 , f (b) = f(a) = \sqrt{9} = 3\\$

After this,

Put the values of a and b to the above equation

$f^i(c) = \frac{f(b) - f(a)}{b - a} \\\\ \frac{1}{{2}\sqrt{c} } = \frac{3 -0}{9-0} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{3}{9} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{1}{3} \\\\ \sqrt[2]{c} = 3\\\\\sqrt{c} = \frac{3}{2} \\\\ c = \frac{9}{4}$

= 2.25

4 0

3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

Which of the following shows the reflection?

AysviL [449]

iii shows the reflection

5 0

3 years ago

Read 2 more answers

Which number does NOT represent a rational number

defon

Answer:

Step-by-step explanation:

B is not rational. B = sqrt(3*3 * 2 * 5) = 3 sqrt(10) sqrt(10) is irrational.

8 0

3 years ago

Read 2 more answers

BRAINLEST + THANKS! (college algebra/ pre-cal)

Jobisdone [24]

The answer is 9m^6/n^8

5 0

3 years ago