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Vikentia [17]
3 years ago
8

What is the measure of angle ADC?

Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0

angle ADC equals 74 degrees

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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:

<u><em>We want to make P subject of the cost function </em></u>

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 = <u><em>mean cost-to-charge ratio for outpatient care</em></u>

\mu_2 = <u><em>mean cost-to-charge ratio for impatient care</em></u>.

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 <u>Two-sample t-test statistics</u> 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, <u><em>the test statistics</em></u>  =  \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.

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

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 <u><em>we reject our null hypothesis</em></u>.

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
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