Solve for 2a:<br> a + 6 = 3a + 2<br> 2a

According to a "how to stop bullying" Web site, 15% of students report experiencing bullying one to three times within the most

AlekseyPX

Answer:

His 95% confidence interval is (0.065, 0.155).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 186, \pi = 0.11$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 - 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.065$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 + 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.155$

His 95% confidence interval is (0.065, 0.155).

4 0

2 years ago

In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

g100num [7]

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

$x^{3}$ = $- 64$

$x^{3} +64$ = 0

We can write 64 as $4^{3}$

$x^{3}$ + $4^{3}$ = 0

using the identity ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

= $(x+4) (x^{2} -x*4+4^{2} )$

= $(x+4)(x^{2} -4x+16)$ ....................(1)

solving the quadratic equation ,

$x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

$x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

<u /> $x=2+-2\sqrt{3i}$ ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

3 0

2 years ago

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

weqwewe [10]

Answer:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

Step-by-step explanation:

1) Data given and notation

n=180 represent the random sample taken

X=125 represent the number of americans between 17 to 24 that not qualify for the military

$\hat p=\frac{125}{180}=0.694$ estimated proportion of americans between 17 to 24 that not qualify for the military

$p_o=0.75$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :

Null hypothesis: $p\geq 0.75$

Alternative hypothesis: $p < 0.75$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

6 0

3 years ago

Solve for c in a+b+c=180

AleksAgata [21]

Answer:

c = -a - b + 180

Step-by-step explanation:

Isolate the c. Note the equal sign, what you do to one side, you do to the other. Subtract a and b from both sides

a (-a) + b (-b) + c = 180 (-a) (-b)

c = 180 - a - b

c = -a - b + 180 is your answer

~

4 0

3 years ago

Select the equation that contains the points (-2 -2) and (4 10)

Brums [2.3K]

Answer: y-6= 2(x-2)

Step-by-step explanation:

find the slope then put it in point slope form.

8 0

2 years ago

Solve for 2a: a + 6 = 3a + 2 2a