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3 years ago

What are some tips/pointers to follow to have the best experiment design possible?

1 answer:

8 0

In order to have a great experiment and great data for analyzing you must have the best experiment design possible.Experiment design is the process of organizing the experiment properly to ensure that the right type and amount of data are available to answer the questions of interest. Some of the tips to follow in order to have good experimental design are:
-Choose a topic that will interest and challenge you.Do not pick a broad topic.
- Spend a lot of time gathering background research.
- Predict the outcome of your experiment with the help of equation.
-Keep a detailed and up-to-date lab notebook with the results.
- Do as many trials as possible .
- During the experiment, do not get discouraged if you run into a lot of problems..

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What is the solution to this system of equations?

Liula [17]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help :))))))))

forsale [732]

Answer:

The answer is A.

Hope that helps :)

Step-by-step explanation:

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3 years ago

1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals n

Andrew [12]

Part 1
We are given $x^2-21x=-4x$ . This can be rewritten as $x^2-18x=0$ .
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}$
$x=\frac{18\pm 18}{2}$

The values of x are
$x_1=\frac{18-18}{2}=0$
$x_2=\frac{18+18}{2}=18$

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.

Part 3.
The first equation is $y=x^2+2$ .
The second equation is $y=3x+20$ .

We have
$x^2+2=3x+20$
$x^2-3x-18=0$
Factoring, we have
$\left(x-6\right)\left(x+3\right)=0$
Equating both factors to zero.
$x_1-6=0\rightarrow x_1=6$
$x_2+3=0\rightarrow x_2=-3$

When the value of x is 6, the value of y is
$y=3\left(6\right)+20=38$

When the value of x is -3, the value of y is
$y=3\left(-3\right)+20=11$

Therefore, the solutions are (6,38) or (-3,11)

7 0

3 years ago

Complete the statement of congruence for the triangles.

Hoochie [10]

If you are proving a congruence of triangles then you need to match the corresponding parts to the triangle you are proving congruent.

BAC would then be congruent to MJR because of the corresponding angles

B is the same angle as M

A is the same angle as J

C is the same angle as R

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3 years ago

Question is attached

AVprozaik [17]

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8 0

3 years ago