What is the measure of angle ADC?

labwork [276]

I believe it is twenty eight. Ten is two thirds of fifteen. Twenty eight is two thirds of forty two. But I'm just assuming tbh

6 0

2 years ago

A box of staples contains 6 x 103 staples and weighs 12 oz. How much does one staple

omeli [17]

Answer:

2 * 10^-3 oz

Step-by-step explanation:

Given that:

Number of staples per box = 6 * 10^3

Weight of staple box = 12 oz

Weight per staple = (weight of staple box / number of staples per box)

Weight per staple = 12 / 6 * 10^3

Weight per staple = 12 / 6 * 1000

= 12 / 6000

= 0.002 oz

= 2 * 10^-3 oz

6 0

2 years ago

Sam has bought a camera for $200 when it was on 20% off sale. What was the original price of the camera

Olenka [21]

Original price = x
x - .2x = 200 ← original price - discount amount = sale price
.8x = 200 ← 1x - 0.2x = .8x
x = 250 ← divided both sides by .8

The original price was $250

5 0

3 years ago

Read 2 more answers

A drive-in theater costs $20 per car, plus an additional fee of $4 per person. The cost, C, of the movie when p people are in th

ozzi

Answer:

$P=\frac{C}{4} -5$

Step-by-step explanation:

Step one:

given

cost per drive= $20

additional fee per person= $4

let the number of persons be p

and the total cost be C

The total cost function is expressed as

$C=20+4p$

Step two:

Required:

We want to make P subject of the cost function

Take 20 to the other side we have

C-20=4p

divide through by 4

P=C-20/4

simplify

P=C/4-5

$P=\frac{C}{4} -5$

8 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

3 years ago