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Angelina_Jolie [31]
2 years ago
15

Xavier is testing different types of planting soil to determine which type is most effective in growing blueberry bushes. He pur

chases two different brands of planting soil from a local store. Xavier applies Brand A to an area of the yard that receives full sunlight and Brand B to another area of the yard that is in partial sunlight. He waters the area with Brand B daily; he waters the area with Brand A every other day. At the end of the study, Xavier concludes Brand A is more effective in growing blueberry bushes. Why is his conclusion not valid?
Mathematics
1 answer:
vladimir1956 [14]2 years ago
4 0

Answer:

He did not control for lurking variables and their impacts on the results of his experiment. Amount of sunlight and water received are two outside variables(or confounding variables) that may impact the growth of his plants and influence the results. He needs to apply the same amount of sunlight and water to each plant within a different planting soil in order to rule out the influence of those two variables and test the sole effect of the soil brand on the plant growth. Otherwise, it would be hard to determine whether his plant growth was because of the soil brand or the different amounts of sunlight and water received

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Determine the area of the given region under the curve <br> (1/x^4) [1,2]
Nikolay [14]

Answer:

\displaystyle \frac{7}{24}

General Formulas and Concepts:

<u>Algebra I</u>

  • Exponential Rule [Rewrite]: \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

[Area] Limits of Riemann's Sums - Integrals

Integration Rule [Reverse Power Rule]:                                                                    \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                          \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle f(x) = \frac{1}{x^4} \\ \ [1, 2]<u />

<u />

<u>Step 2: Find Area</u>

  1. [Integral] Set up area:                                                                                    \displaystyle \int\limits^2_1 {\frac{1}{x^4}} \, dx
  2. [Integral] Rewrite:                                                                                            \displaystyle \int\limits^2_1 {x^{-4}} \, dx
  3. [Integral] Reverse Power Rule:                                                                      \displaystyle \frac{-1}{3x^3} \bigg| \limits^2_1
  4. [Area] Fundamental Theorem of Calculus:                                                   \displaystyle \frac{7}{24}

Topic: Calculus

Unit: Basic Integration/Riemann Sums

Book: College Calculus 10e

6 0
2 years ago
what does y =mx +b mean and how do you use it to solve problems? whats the different between y=mx+b and y=mx + c
algol13

y=mx+c is the equation for a line graph, m is the gradient and c is where it goes through the y axis, the only difference between y=mx+c and y=mx+b is where u live as in different countries learn it different ways but it is still the same thing
4 0
3 years ago
The distance AB rounded to the nearest tenth = [ ? ]
docker41 [41]

Answer:

4.5 units

Step-by-step explanation:

To get the distance we need the coordinates of the points

A is ( 3,1)

B is (-1,-1)

Let A be (x1,y1) and B be (x2,y2)

So applying the formula, we have the distance as;

D = √(-1-3)^2 + (-1-1)^2

D = √(16 + 4)

D = √20

D = 4.5 units

7 0
3 years ago
HELPPP !!!
Masja [62]
I just realized I forgot a whole unit we did this year, but a(n)=5n-12
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7 0
3 years ago
Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
3 years ago
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