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3 years ago

My stoopid ahh needs help with something else someone HELP

Mathematics

2 answers:

Pepsi [2]3 years ago

7 0

B because you don’t know exactly how many times she’s gonna grab the marbles

gregori [183]3 years ago

6 0

B is the correct answer

You might be interested in

Pairs that satisfy function y=2x+1

evablogger [386]

Test a pair from each table by substituting their values into the given equation and solving.

A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1

B) y = 2x + 1 .......... 1 = 2(0) + 1 .......... 1 = 1

C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1

D) y = 2x + 1 .......... -2 = 2(0) + 1 .......... -2 ≠ 1

The only pair that satisfied the equation was from answer choice B. Therefore, B is the correct answer.

8 0

3 years ago

Choose which property of equality is demonstrated moving from step a to step b. a. If b = 3, and m = 5, and y = mx + b then, b.

Step2247 [10]

Answer:

add the = problome

Step-by-step explanation:

5 0

3 years ago

Find C. to form completing Square <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-%5Cfrac%7B3%7D%7B2%7Dx%20%2B%20C" id="Tex

nata0808 [166]

Answer:

c = 9/16

Step-by-step explanation:

(a-b)²

a² - 2ab + b²

here a = x

b = ?

x² - (2)(x)(?) + ?²

Now we can see that -(2x)? = -3x/2

-2x? = -3x/2

cancel -x on both sides

2? = 3/2

? = 3/2 * (2)

? = 3/4

so b = 3/4

(x² - 3x + 9/16) = (x-3/4)²

so from the expression we can see that C = 9/16

3 0

4 years ago

You and 19 other people are entered in a drawing. One person will be randomly selected to win a prize. What is the probability t

Novosadov [1.4K]

Answer:

Step-by-step explanation:

1 person will be selected out of a total of 20 (you + 19)

The probability is 1/20 = 0.05 = 5%

4 0

4 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