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3 years ago

What's the purpose of life? why do we exist? are aliens real?

Mathematics

2 answers:

Solnce55 [7]3 years ago

7 0

Answer: we exist to discover things. their are many opinion but it is good to understand aliens are real.

frez [133]3 years ago

3 0

I personally beleive that we are meant to be god's creation to have our race keep going and love so we can keep reproducing, so we can live on. :)

You might be interested in

A small publishing company is planning to publish a new book. The production costs will include one-time fixed costs (such as ed

Blizzard [7]

Answer: 2,435

Step-by-step explanation:

trust

3 0

3 years ago

Kyla works as a receptionist. Last week she made $236.40 for working 24 hours. About how many hours would she need to work to ea

kow [346]

Answer:

32.99

Step-by-step explanation:(you might have to round) if you have to round it or.. but this is it because

236.40 divided by 24 is 9.85 so

325 divided by 9.85 is 32.99 (you only need 2 decimal places for these kinds of questions because of repeating decimals).

Hope this helps :)

8 0

3 years ago

Write an equation of a line that is parallel to the given line.

Marina CMI [18]

Answer:

5) y=5

6)x=7

7)a=18

Step-by-step explanation:

5) y= any number will work. don't add a variable like x

6)x= any number will work. don't add a variable

My work for 7 is in the picture. If you'd like me to explain more for 7 lmk :)

6 0

3 years ago

Write the equation of the line in slope-intercept form that passes through

never [62]

Answer:

y= -2/3x+3

Step-by-step explanation:

Parallel lines have the same slope but different y-intercepts.

y=-2/3x is your original equation

Your slope is still -2/3

y=mx+b

y=-2/3x + b

Substitute the points from the coordinates

5 = -2/3(-3) + b

5 = 2 + b

-2 -2

b=3

Your parallel line equation is: y= -2/3x+3

Have a great day!

3 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