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kupik [55]
2 years ago
8

By definition, when a test or technique measures what it intends to, it is a. valid. b. reliable. c. significant. d. generalizab

le.
Mathematics
1 answer:
SOVA2 [1]2 years ago
7 0

By definition, when a test or technique measures what it intends to, it is valid.

The answer is option A.

The criterion-related validity of a take-a-look is measured by the validity coefficient. Its miles said as a number between zero and 1.00 that indicates the significance of the relationship, "r," among the take a look at and a measure of task overall performance (criterion).

Reliability and validity are both approximately how nicely a technique measures something: Reliability refers to the consistency of a measure (whether or not the consequences can be reproduced below identical situations).

Validity refers back to the accuracy of a measure (whether or not the results really do constitute what they may be speculated to a degree).

Learn more about Validity here: brainly.com/question/25759088

#SPJ4

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Which of the following is equal to the expression below?
just olya [345]

\frac{\sqrt{9}}{\sqrt{81}}=\frac{3}{9}=\boxed{\frac{1}{3}}

The answers are A and B.

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The difference of a number x and 9 is fewer then 4
Sav [38]
The equation would be:
x - 9 < 4
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8 0
3 years ago
Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 &lt; 2pi. Show work please!
yan [13]

Answer:

\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}

Step-by-step explanation:

You need 2 things in order to solve this equation:  a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

cos(2\theta)=1-2sin^2(\theta)

so we will make that replacement, getting everything in terms of sin:

sin(\theta)+1=1-2sin^2(\theta)

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

2sin^2(\theta)+sin(\theta)=0

We can factor out the sin(theta), since it's common in both terms:

sin(\theta)(2sin(\theta)+1)=0

Because of the Zero Product Property, either

sin(\theta)=0 or

2sin(\theta)+1=0

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi.  They are:

\theta=\frac{\pi}{2},\frac{3\pi}{2}

The next equation needs to first be solved for sin(theta):

2sin(\theta)+1=0 so

2sin(\theta)=-1 and

sin(\theta)=-\frac{1}{2}

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval.  They are:

\theta=\frac{2\pi}{3},\frac{4\pi}{3}

7 0
3 years ago
D\dx(3x^2)<br> Please answer fast.
Sloan [31]

Answer:

3x

Step-by-step explanation:

\frac{d}{dx} (3x  {}^{2} )

\frac{1}{x} 3 {x}^{2}

3x

7 0
3 years ago
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