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3 years ago

Find y for x= 1,2,3,and 4y=2x y=3x+1 y=x-5 y=-5x+6

1 answer:

8 0

Like I said, just use the formulas
y=2x is 2,4,6,8
y=3x+1 is 4,7,10,13
y=x-5 is -4,-3,-2,-1
y=5x+6 is 11,16,21,26
Wow I just noticed these are all patterns...

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A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

3 0

3 years ago

Hypotheses: p=.5 vs p NE .5 ; Sample data: p=28/40= .70 with n=40 Round the p-value to three decimal places.

jeka57 [31]

Answer:

hahahahahaha yes hnjoo

7 0

3 years ago

I have $100. I spent $40 on a prize and $45 on food. How do I figure out what % is purchase is from the original amount I had?

Paladinen [302]

You could say that you spent (purchased) 85% of the money that you had.

You spent a total of $85.

If you create the fraction 85/100, that is the same as saying you spent 85% of the money.

8 0

3 years ago

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Vika [28.1K]

$\longrightarrow{\mathfrak{\frac{-9}{6}÷\frac{3}{-2}}}$