Which is true about the completely simplified difference of the polynomials 6x6 − x3y4 − 5xy5 and 4x5y + 2x3y4 + 5xy5?

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10

son4ous [18]

There is a relationship between confidence interval and standard deviation:
$\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}$
Where $\overline{x}$ is the mean, $\sigma$ is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first:
$\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}$
When we do all the calculations we find that:
$\overline{x}=123.8\\ \sigma=11.84$
Now we can find confidence intervals:
$($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\$
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing

6 0

2 years ago

Find the quantity q that maximizes profit if the total revenue, R(q), and total cost, C(q) are given in dollars by R(q)=3q−0.002

marysya [2.9K]

Answer:

The value of q that maximize the profit is q=200 units

Step-by-step explanation:

we know that

The profit is equal to the revenue minus the cost

we have

$R(q)=3q-0.002q^{2}$ ---> the revenue