9. Which of the following is the interquartile range of the data set shown below?

What is the rate of change of equation x=9

Delicious77 [7]

Erm..

9 I guess..

Good luck?

5 0

3 years ago

Help me please...!!!!!!

goblinko [34]

B would be the answer for sure

4 0

3 years ago

Read 2 more answers

Enter the values to complete the statement. To determine whether (−1,4) (−1,4) is a solution to the equation 3x+8y=29 3x+8y=29 ,

aivan3 [116]

To determine whether (−1,4) is a solution to the equation 3x+8y=29, substitute ...-1... for x and ...4... and for y.

In a pair of values like (-1, 4), the first coordinate, or entry, occupied by -1 always represents x.

Similarly, the second coordinate occupied by 4 represents y.

Answer: substitute ...-1... for x and ...4... and for y.

8 0

3 years ago

What will happenif you: multiply one charge by 2.0, multiply the other charge by 4.8, and multiply the distance between them by

noname [10]

Yes it is.
F1 = k q1 * q2 / r^2
F2 = k *(2*q1) * (4.8 q2) / (7.2 r^2)

Work on F2 for a moment.
F2 = k * 9.6 (q1*q2) / 51.84
F2 = (9.6/ 51.84) * k * q1 * q2 / r^2
Since k * q1 * q2/r^2 is the same in both questions let kq1*q2/r^2 = m

F1 = m
F2 = 0.18 m

So to make it easier F2/F1 = 0.18m / m
F2/ F1 = 0.18 the m's cancel.

And that should be how you do the question.

4 0

3 years ago

The Oregon Department of Health web site provides information on the cost-to-charge ratio (the percentage of billed charges that

Alekssandra [29.7K]

Answer:

We conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

Step-by-step explanation:W

We are given with the cost-to-charge ratios for both inpatient and outpatient care in 2002 for a sample of six hospitals in Oregon below;

Hospital 2002 Inpatient Ratio 2002 Outpatient Ratio

1 68 54

2 100 75

3 71 53

4 74 56

5 100 74

6 83 71

Let $\mu_1$ = mean cost-to-charge ratio for outpatient care

$\mu_2$ = mean cost-to-charge ratio for impatient care.

SO, Null Hypothesis, $H_0$ : $\mu_1 \geq \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}

Alternate Hypothesis, $H_A$ : $\mu_1 < \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care}

The test statistics that would be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1_+_n_2_-_2$

where, $\bar X_1$ = sample mean cost-to-charge Outpatient Ratio = $\frac{\sum X_1}{n_1}$ = 63.83

$\bar X_2$ = sample mean cost-to-charge Impatient Ratio = $\frac{\sum X_2}{n_2}$ = 82.67

$s_1$ = sample standard deviation for Outpatient Ratio = $\sqrt{\frac{\sum (X_1-\bar X_1 )^{2} }{n_1-1} }$ = 10.53

$s_2$ = sample standard deviation for Impatient Ratio = $\sqrt{\frac{\sum (X_2-\bar X_2 )^{2} }{n_2-1} }$ = 14.33

$n_1$ = sample of hospital for outpatient care = 6

$n_2$ = sample of hospital for outpatient care = 6

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(6-1)\times 10.53^{2}+(6-1)\times 14.33^{2} }{6+6-2} }$ = 12.574

So, the test statistics = $\frac{(63.83-82.67)-(0)}{12.574 \times \sqrt{\frac{1}{6}+\frac{1}{6} } }$ ~ $t_1_0$

= -2.595

The value of t test statistics is -2.595.

Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.

Since, our test statistics is less than the critical value of t as -2.595 < -1.812, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

8 0

4 years ago