Consider the function f(x) = x^{2} on the interval [1, 9]. Let P be a uniform partition of [1,9] with 16 sub-intervals. Compute the left and right Riemann sum of f on the partition. Use exact values.

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: ${\displaystyle 0.127<p<0.191}$. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___

In a science fair​ project, Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left​ hand, and then she asked the therapists to identify the selected hand by placing their hand just under​ Emily's hand without seeing it and without touching it. Among 358 ​trials, the touch therapists were correct 172 times. Complete parts​ (a) through​ (d).
a) Given that Emily used a coin toss to select either her right hand or her left​ hand, what proportion of correct responses would be expected if the touch therapists made random​ guesses? ​(Type an integer or a decimal. Do not​ round.)
b) Using​ Emily's sample​ results, what is the best point estimate of the​ therapists' success​ rate? ​(Round to three decimal places as​ needed.)
c) Using​ Emily's sample​ results, construct a ${\displaystyle 90\mathrm{\%}}$ confidence interval estimate of the proportion of correct responses made by touch therapists.
Round to three decimal places as​ needed - ?${\displaystyle <p<}$?

Find a 95 confidence interval for ${\displaystyle \theta }$ based on inverting the test statistic statistic ${\displaystyle \hat{\theta }}$.
For our data we have ${\displaystyle Y_i\sim N(\theta x_i,1)\text{for}i=1,\dots ,n.}$
Therefore it can be proven that the MLE for ${\displaystyle \theta }$ is given by
${\displaystyle \hat{\theta }=\frac{\sum x_i{Y}_i}{\sum x_i^2}}$
To find the confidence interval I should invert the test statistic ${\displaystyle \hat{\theta }}$.
The most powerful unbiased size ${\displaystyle \alpha =0,05}$ test for testing
${\displaystyle H_0:\mu ={\mu }_0\text{vs.}{H}_1:\mu \ne {\mu }_0}$
where ${\displaystyle X_1,\dots ,X_n\sim \text{ iid }n(\mu ,{\sigma }^2)}$ has acceptance region
$A\left({\mu }_0\right)=\left\{\mathbf{x}:|\overline{x}-{\mu }_0|\le 1,96\sigma /\sqrt{n}\right\}.$
Substituting my problem (I think) we get that the most powerful unbiased size ${\displaystyle \alpha =0,05}$ test for testing
${\displaystyle H_0:\theta =\hat{\theta }\text{vs.}{H}_1:\theta \ne \hat{\theta }}$
has acceptance region $\{A\left(\hat{\theta }\right)=\{\mathbf{y}:|\bar{y}-\hat{\theta }|\le 1,\frac{96}{\sqrt{n}}\}$
or equivalently, $A\left(\hat{\theta }\right)=\left\{\mathbf{y}:\frac{\sqrt{n}\overline{y}-1,96}{\sqrt{n}{x}_i}\le \hat{\theta }\le \frac{\sqrt{n}\overline{y}+1,96}{\sqrt{n}{x}_i}\right\}$
Substituting ${\displaystyle \hat{\theta }=\sum x_{i}Y\frac{i}{\sum }{x}_i^2}$ we obtain
$A\left(\hat{\theta }\right)=\left\{\mathbf{y}:\frac{\sqrt{n}\overline{y}-1,96}{\sqrt{n}{x}_i}\le \frac{\Sigma x_{i}Yi}{\Sigma x_i^2}\le \frac{\sqrt{n}\overline{y}+1,96}{\sqrt{n}{x}_i}\right\}$
This means that my ${\displaystyle 1-0,05=0,95\left(95\mathrm{\%}\right)}$ confindence interval is defined to be
${\displaystyle C\left(y\right)=\{\hat{\theta }:y\in A\left(\hat{\theta }\right)\}}$
But I can't seem to find anything concrete and I feel that I've made mistakes somewhere. What to do?