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dalvyx [7]
4 years ago
11

The awser to (18)⋅(3)=

Mathematics
1 answer:
qwelly [4]4 years ago
7 0

Answer:

108 I think

Hope this helps :)

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Solve this answer for my test thank you
Naddik [55]

Answer:

A

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3 years ago
A binomial random variable is what type of random variable?
OLga [1]

Answer:

It is a discrete random variable.

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Counts often when something happens like in fixed numbers of tries.

Hope I helped!:)

4 0
3 years ago
Hurry please!!!! I can’t decide
sesenic [268]

Answer:

Change 2 to 3 and the last one will be 2

Step-by-step explanation:

I believe a triangular prism will be 1/2 instead of 1/3 because of the side differences between a triangular prism and a Pyramid

Hope this Helped

4 0
3 years ago
SOME ONE HELP ME I REALLY NEED A SOLUTION FOR THIS PROBLEM WITH A CLEAR EXPLANATION
Sergio [31]

I'm going to answer this question using logic.  Let x be Jenny's favorite number.  We are going to square root this number, \sqrt{x} and then multiply it by \sqrt{2}.

This product needs to be an integer.  How is that obtainable? To be an integer, we need to get rid of the nasty \sqrt{2} part.  The only way I can think of to get rid of it, is to multiply it by \sqrt{2}, because \sqrt{2} *\sqrt{2} = 2.  Thus we have some conditions we need to fulfill when choosing Jenny's favorite number.

When we take the square root of Jenny's favorite number, x, it must contain a perfect square and a 2 in its prime factorization.  For example, 8 works because 8 = 2 x 2 x 2, or 2² x 2.  

You notice 8 is made up of a perfect square multiplied by 2.  So when we take the square root of 8, we get:

\sqrt{8} = \sqrt{2^{2}*2 } = \sqrt{ 2^{2} }*\sqrt{2} = \sqrt{4} *\sqrt{2} = 2*\sqrt{2}

So 8 is the same thing as 2\sqrt{2}

So when we multiply this by \sqrt{2}, we will get an integer! So as long as Jenny's favorite number consists of a perfect square and two in its prime factorization, we will have an integer!

So possible choices are: 2,8,18.

Why does 18 work? Because \sqrt{18} = \sqrt{9}* \sqrt{2} = 3\sqrt{2}

When we multiply this by \sqrt{2}, we get 6, which is an integer.

b) Suppose instead of multiplying by \sqrt{2}, we divided by \sqrt{2}. Is the resulting quotient still an integer?

YES, because we can get rid of the \sqrt{2} part by dividing by \sqrt{2} as well.  This leaves only the "perfect square" part left in our square root, and obviously a perfect square is an integer when we square root it.

I hope that made sense! (⌐■_■)

3 0
3 years ago
PLZZ HELLLLLPPPPievjp
RUDIKE [14]
She can ride a bike 7 miles per hour
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3 years ago
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