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kenny6666 [7]
3 years ago
11

During an experiment, readings for blood pressure in a person's body were found to be constant. However, when measured by a diff

erent blood pressure cuff, the readings differed by 15 points for each reading. This difference indicates that the results are (independent,reliable,valid) but not (dependent,reliable,valid).

Mathematics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

the difference indicates that the results are independent but not reliable

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A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, gi
defon

Answer:

An ordered pair is a set of inputs and outputs and represents a relationship between the two values. A relation is a set of inputs and outputs, and a function is a relation with one output for each input.

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here is a exstended explantion:

A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, given an element x in X, there is only one element in Y that x is related to.

For example, consider the following sets X and Y. I'll give you a relation between them that is not a function, and one that is.

   X = { 1, 2, 3 }

   Y = { a , b , c, d }

   Relation from X to Y (i.e., in XxY ) : { (1,a) , (2, b) , (2, c) , (3, d) }

This relation is not a function from X to Y because the element 2 in X is related to two different elements, b and c. (Note, if you transpose the ordered pairs you <i>would</i> have a function from Y to X - can you see WHY?)

   Relation from X to Y that is a function: { (1,d) , (2,d) , (3, a) }

This is a function since each element from X is related to only one element in Y. Note that it is okay for two different elements in X to be related to the same element in Y. It's still a function, it's just not a one-to-one function.

8 0
3 years ago
Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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Hypotenuse= \sqrt{1.5^{2}+ 4^{2}  }
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Next year Phil and Matt may sell the cookies for $50 each. They plan to make the same total number of cookies, but they predict
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No, they should not raise the price of the cookies. Because they are going to sell more base on the equation above if they stick with the original price.

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