Answer:
A
Step-by-step explanation:
it is going from low to high
Answer:
15 feet of water spread
Step-by-step explanation:
Area of the circular pattern is 706.50 sq. feet
Area = A = π r^2 = (3.14)* r^2
706.50 = (3.14)* r^2
r^2 = 706.50 / 3.14
r^2 = 706.50 / 3.14
r = root (225)
r = 15 feet
Answer:

General Formulas and Concepts:
<u>Calculus</u>
Integration
Integration Rule [Reverse Power Rule]: 
Integration Rule [Fundamental Theorem of Calculus 1]: 
Integration Property [Multiplied Constant]: 
Integration Property [Addition/Subtraction]: ![\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7B%5Bf%28x%29%20%5Cpm%20g%28x%29%5D%7D%20%5C%2C%20dx%20%3D%20%5Cint%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%5Cpm%20%5Cint%20%7Bg%28x%29%7D%20%5C%2C%20dx)
Area of a Region Formula: ![\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%20%3D%20%5Cint%5Climits%5Eb_a%20%7B%5Bf%28x%29%20-%20g%28x%29%5D%7D%20%5C%2C%20dx)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
y = 2x + 3
<em>x</em>-interval [3, 4]
<em>x</em>-axis
<em>See attachment for graph.</em>
<u>Step 2: Find Area</u>
- Substitute in variables [Area of a Region Formula]:

- [Integral] Rewrite [Integration Property - Addition/Subtraction]:

- [Integrals] Rewrite [Integration Property - Multiplied Constant]:

- [Integrals] Integrate [Integration Rule - Reverse Power Rule]:

- [Integrals] Integrate [Integration Rule - FTC 1]:

- Simplify:

∴ the area bounded by the region y = 2x + 3, x-axis, and the coordinates x = 3 and x = 4 is equal to 10.
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Learn more about integration: brainly.com/question/26401241
Learn more about calculus: brainly.com/question/20197752
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
Step-by-step explanation:
Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
This is because at a higher level of discrepancies, there will be a strong evidence against the null. This means that it will be rare to find discrepancies if null was true.
In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.