HELLPPPPPP please it’s due today

Find the equation of a line in slope-intercept form, containing the points (9,2.6) and (8,2.2).

Zolol [24]

The equation in slope-intercept form would be y = 0.4x - 1

hope this helps!!

4 0

3 years ago

Several years ago, 50% of parents who had children in grades K-12 were satisfied with the quality of education the students r

galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

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:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612$

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

5 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$