Can you solve this I am confused

Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0 = 2x2 + 3x – 8? Quadratic formula: x

Lynna [10]

ANSWER

1.39

EXPLANATION

The given quadratic equation is

$0 = 2 {x}^{2} + 3x - 8$

This is the same as,

$2 {x}^{2} + 3x - 8 = 0$

Comparing to

$a {x}^{2} + bx + c = 0$

We have

a=2, b=3,c=-8

Using the quadratic formula, the solution is given by:

$x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute the values to get,

$x = \frac{ - 3\pm \: \sqrt{ {3}^{2} - 4(2)( - 8)} }{2(2)}$

$x = \frac{ - 3\pm \: \sqrt{ 73} }{4}$

The positive root is

$x = \frac{ - 3 + \: \sqrt{ 73} }{4} = 1.39$

to the nearest hundredth.

4 0

3 years ago

Read 2 more answers

HELP ME PLZ I WILL MARK LCM

Nataly [62]

Answer:

18

Step-by-step explanation:

Find the multiples of 9

9,18,27,36

Find the multiples of 6

6,12,18,24

The first common multiple is 18

8 0

3 years ago

Read 2 more answers

Determine the value of x in the diagram

Sidana [21]

Answer:

3x + x + 3x + x = 360

8x = 360

x = 45

3(45)= 135

Step-by-step explanation:

3 0

3 years ago

Through: (-4,-3), slope = 1<br> Standard form from a point and a slope

morpeh [17]

Answer:

y=-4/-3x+1

Step-by-step explanation:

8 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago