Evaluate the product of 12 and 2 minus the quotient of 45 and 5

The average rate of change of f from 2 to 3

ella [17]

For a function of $f$ , the average rate of change can be found as follows:

$ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$

So, this problem asks for the Average Rate of Change of $f$ from $x_{1}=2$ to $x_{2}=3$ . In this way, we need to find $f(x_{1}) \ and \ f(x_{2})$ . As you can see above, we have the graph of $f(x)$ , so we can find these values. Thus, from the graph:

$f(x_{1})=4 \\ f(x_{2})=1$

Therefore:

$ARC=\frac{1-4}{3-2} \\ \\ \therefore \boxed{ARC=-3}$

3 0

3 years ago

The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and

kykrilka [37]

Answer:

The calculated χ² = 17.0281 falls in the critical region χ² ≥ 9.49 so we reject the null hypothesis that the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)= 4 d.f

4) Computations:

Under H0 ,

Observed Expected E χ²= ∑(O-e)²/e

40 35.00 0.71

25 24.50 0.01

5 10.50 2.88

35 40.00 0.62

35 28.0 1.75

10 12.00 0.33

25 25.00 0.00

10 17.50 3.21

15 7.50 7.50

∑ 17.0281

Column Totals 100 70 30 200 (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² = 17.0281 falls in the critical region χ² ≥ 9.49 so we reject the null hypothesis that the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0

3 years ago

An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline

vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have a seat, we have

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 = 5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

$z = \frac{x-np}{\sqrt{np(1-p)} } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} }$ = - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = $\sqrt{npq}$ = 2.233

P (x≤100) = P $(z$ = P(z<0.34) = 0.63307.

6 0

3 years ago

Which statement describes the system of equations

Contact [7]

Answer:

2y-2x-2y= - 22x = 0x + 2y = 1

Step-by-step explanation:

7 0

3 years ago

Determine (a) the slope of the line parallel to the line whose slope is m=2 and (b) the slope of the line perpendicular to the l

kotykmax [81]

Answer:

(b) $\displaystyle -\frac{1}{2} = m$

(a) $\displaystyle 2 = m$

Step-by-step explanation:

Perpendicular Lines have OPPOSITE MULTIPLICATIVE INVERSE RATE OF CHANGES [SLOPES], whereas Parallel Lines have SIMILAR RATE OF CHANGES [SLOPES].

I am joyous to assist you anytime.

7 0

3 years ago