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Sladkaya [172]
2 years ago
13

Isprobibc) 23 001 - 22 999​

Mathematics
1 answer:
ZanzabumX [31]2 years ago
7 0

Answer

I don't know

Step-by-step explanation:

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What Two pairs of fractions are shown. Which symbol correctly compares 2/3 2/4 each pair of fractions?
VikaD [51]

Answer:

Greater than symbol.

2/3 > 2/4

Step-by-step explanation:

We have two fractions, 2/3 and 2/4

We then go back to the inequality part of mathematics where we have symbols such as greater than (>), less than (<), and equal to (=).

Of the three inequality symbols, we look and pick at the one that represents best, the fractions we were given.

2/3 and 2/4

With 2/3 being greater than 2/4.

2/3 = 0.67

2/4 = 0.50

Succinctly, 2/3 > 2/4.

4 0
2 years ago
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
Which is the equation of the given line in slope intercept form (-1, 2) (1, -4)
Degger [83]

Answer:

y = -3x - 1

Step-by-step explanation:

y = mx + c

m = (-4-2)/(1-(-1)) = -6/2 = -3

y = -3x + c

When x = -1, y = 2

2 = -3(-1) + c

c = 2 - 3 = -1

y = -3x - 1

5 0
3 years ago
Pls answer my question it is urgent
yaroslaw [1]

Answer: rs 400 and 500

Step-by-step explanation:

3 0
2 years ago
Darius is a graphic designer he makes a sign for a company that has a width of 2 feet and a length of 2 feet 6 inches the compan
Sav [38]

Answer:

\frac{1}{3}

Step-by-step explanation:

Width of the sign made by Darius initially = 2 feet

Length of the sign = 2 feet 6 inches

Dimension of the flyer are 10 inches by 8 inches. Since, in original sign the measure of length is greater, therefore, the length of flyer is 10 inches and its width is 8 inches.

In order to find the scale factor we must convert the lengths and widths to same units. Lets convert the length and width of sign into inches.

Since, 1 feet = 12 inches

Length of the sign = 2 feet 6 inches = 2(12) + 6 inches = 30 inches

Width of the sign = 2 feet = 2(12) inches = 24 inches

Now we can find the scale factor by either comparing the lengths or widths of both the designs. Scale factor will be equal to the ratio of corresponding lengths/widths.

So, the scale factor would be:

Length of Flyer : Length of Sign

= 10 inches : 30 inches

= 1 : 3

= \frac{1}{3}

This shows, the length of flyer is \frac{1}{3} times as that of the sign. So, the scale factor that Darius must use is \frac{1}{3}. The length and width of the flyer are \frac{1}{3} as that of the sign.

The same scale factor would result if we would have used the ratio of widths instead of the lengths.

Scale Factor = Width of Flyer : Width of sign

= 8 : 24

= 1 : 3

Therefore, Darius must type the scale factor of \frac{1}{3} in his computer to get the size of the flyer.

7 0
2 years ago
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