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3 years ago

-9x + 6 = -x + 4 help pls ill give brainliest

Mathematics

2 answers:

tatiyna3 years ago

7 0

Answer: x = 1/4

Step-by-step explanation:

-9x + 6 = -x + 4

+x +x

-8x + 6 = 4

-6 -6

-8x = -2

divide by -8 on both sides

x = 2/8

simplify

x = 1/4

garik1379 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

-9x + 6 = -x + 4

-9x + x = 4 - 6

-8x = -2

x = 2/8

x = 1/4

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Does 6/4 = 1 2/4? I need awnsers by tmrw

Gala2k [10]

Answer:

Yes

Step-by-step explanation:

There is definetly a whole number so thats correct

1. subtract 4 from 6 equals 2

2. turn that 4 to a 1 whole and leave what's alone

3. Look at what you did... 1 2/4

4. Take pride

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3 years ago

2(1/2q+1)=-3(2q-1)+4(2q+1)

babymother [125]

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3 years ago

What must be done to sentence 3 ?

neonofarm [45]

Correct answer is B. Begin it with the words "Antarctica is".

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6 0

2 years ago

Marjorie wrote the fraction 7/10 as a fraction with a denominator of 100. She wrote the fraction as a decimal. Find and correct

Tatiana [17]

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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