How many minutes in 255 seconds A)2 11/12 B)3 5/12 C)3 3/4 D) 4 1/4

Solve the equation for x, where x is a real number (5 points): -2x^2 + 5x + 2 = -3

Juli2301 [7.4K]

Add 3 to both sides so that the equation becomes -2x^2 + 5x + 5 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                              x = [ -b ± √(b^2 - 4ac) ] / (2a)
                              x = [ -5 ± √(5^2 - 4(-2)(5)) ] / ( 2(-2) )
                              x = [-5 ± √(25 - (-40) ) ] / ( -4 )
                              x = [-5 ± √(65) ] / ( -4)
                              x = [-5 ± sqrt(65) ] / ( -4 )
                              x = 5/4 ± -sqrt(65)/4
The answers are 5/4 + sqrt(65)/4 and 5/4 - sqrt(65)/4..

5 0

3 years ago

Dan invests £8000 into his bank account.

Serhud [2]

If 8000 dollars multiplied by 2% is 160 dollars and 160 dollars multiplied by 3 years gives you the answer of 480 dollars

6 0

3 years ago

Read 2 more answers

three brothers and a sister shared a sum of money equally among themselves. if the brothers alone had shared the money, then the

Virty [35]

Answer:

1. t=240

2.t/3=(t/4)+20

Step-by-step explanation:

4 0

3 years ago

An adult takes 800 mg of ibuprofen. Each hour, the amount of ibuprofen in the person's

ss7ja [257]

Answer:

116.31468544 mg ibuprofen left

Step-by-step explanation:

Since there are 800 mg at the beginning and the system decreases by about 32% of the ibuprofen

After 1 hour

→ 800 - (800x32/100) = 800 - 256 = 544 mg ibuprofen left

After 2 hours

→ 544 - (544x32/100) = 544 - 174.08 = 369.92 mg ibuprofen left

After 3 hours

→ 369.92 - (369.92x32/100) = 369.92 - 118.3744 = 251.5456 mg ibuprofen left

After 4 hours

→ 251.5456 - (251.5456x32/100) = 251.5456 - 80.494592 = 171.051008 mg ibuprofen left

After 5 hours

→ 171.051008 - (171.051008x32/100) = 171.051008 - 54.73632256 = 116.31468544 mg ibuprofen left

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