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3 years ago

Find th area of the circle. Use 3.14 for π round to the nearest unit.

Mathematics

2 answers:

Romashka [77]3 years ago

7 0

Answer:

the answer is c. i have done this quiz already

kramer3 years ago

4 0

I think the answer is C because you would A=Pi x r x 2

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6*4 equals 24 bam easy mat

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3 years ago

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Which one is the unit of force A.Joules B. Newton C. Celsius D. Kelvin

Snezhnost [94]

Answer:

Newtons

Step-by-step explanation:

Joules measure electrical power and celcius and Kelvin measures temperature.

7 0

3 years ago

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The radius of a semicircle is 3 feet. What is the semicircle's area? Use 3.14 for Pi

valentinak56 [21]

Answer:

14.13 ft²

Step-by-step explanation:

The area of a semicircle is

$= \frac{\pi \: {r}^{2} }{2}$

We were given the radius of the semicircle to be: r=3 feet.

We were also asked to use π as 3.14.

We substitute into the formula to get:

$= \frac{3.14 \times {3}^{2} }{2}$

This is the same as:

$= \frac{3.14 \times 9}{2}$

$= 3.14 \times 4.5$

$= 14 .13 {ft}^{2}$

The area of the semicircle is 14.13 ft²

5 0

2 years ago

Can someone explain how to do this?

Over [174]

Answer:

Step-by-step explanation:

The two legs of an iscosles right triangle are equal. The sides opposite each 45 degree angle are equal.

6^2 + 6^2 = x^2

36 + 36 = x^2

x^2 = 72

sqrt(x^2) = sqrt(72)

x = 6√2

3 0

2 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