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The median rainfall during a spring storm in the lowlands is about 4 mm less than a spring storm in the highlands. The mean rain

mariarad [96]

There is a difference because in the spring, there may be more days of rain, therefore that would increase the mean for the spring compared to the other days.

5 0

3 years ago

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Danny opened a savings account with $50 and depositedn$10 each week for 4 weeks

Romashka [77]

it is 40$$$Answer:

Step-by-step explanation:

3 0

3 years ago

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Find the distance from point C to point D.

sineoko [7]

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. The distance from point C to point D is 258.8 meters.

<h3>What is Tangent (Tanθ)?</h3>

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. it is given as,

$\rm Tangent(\theta) = \dfrac{Perpendicular}{Base}$

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

The distance from point C to D is equal to the distance from point A to B. Therefore, the height of the building can be written as,

Tan(25°) = AB / BC

Tan(25°) × 555 m = AB

AB = 258.8 meters

Hence, the distance from point C to point D is 258.8 meters.

Learn more about Tangent (Tanθ):

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3 0

2 years ago

Y=–3x + 6y=9 What is the solution to the system of equations? (–21, 9) (9, –21) (–1, 9) (9, –1)

weqwewe [10]

ANSWER

The solution is $(-1,9)$

EXPLANATION

We want to solve the simultaneous equations

$y=-3x+6---(1)$

and

$y=9--(2)$ .

We substitute equation (2) in to equation (1), to obtain

$9=-3x+6$

This has now become a linear equation in a single variable $x$ .

We solve for x by grouping like terms.

$9-6=-3x$

$3=-3x$

We divide through by negative 3 to get;

$-1=x$ .

Hence, the solution is $(-1,9)$

7 0

3 years ago

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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