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mihalych1998 [28]
2 years ago
12

A probability experiment will consist of selecting one student at random from your math class. Is the probability that the selec

ted student is female greater than, equal to,or less than the probability that the selected student is male? Explain your reasoning

Mathematics
1 answer:
kumpel [21]2 years ago
8 0
Not likely because in my class we have more boys then girls 
m=27
f=5
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Use complete sentences to describe why √-1 ≠ -√1
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Well let's say that to compare these two numbers, we have to start with the definition first.

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\displaystyle \large{ {y}^{2}  = x} \\  \displaystyle \large{ y =  \pm  \sqrt{x} }

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

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\displaystyle \large{  {a}^{2}  = a \times a =  |b| }

Where b is the result from a×a. Let's see an example.

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So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

\displaystyle \large{   {y}^{2}  =  - 1}

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

\displaystyle \large{   \sqrt{ {y}^{2} } =   \sqrt{ - 1}  }

Then where does plus-minus come from? It comes from one of Absolute Value propety.

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Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

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<u>C</u><u>o</u><u>n</u><u>c</u><u>l</u><u>u</u><u>s</u><u>i</u><u>o</u><u>n</u>

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

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