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Ghella [55]
3 years ago
7

Find the number of positive integers n is less than or equal to 1000 such that 15n is a perfect square.

Mathematics
1 answer:
boyakko [2]3 years ago
3 0

Answer:

8 positive integers.

Step-by-step explanation:

One value of n would be 15 because 225 = 15^2.

Other values  are 225 * n where 15n  <= 1000 and n is a perfect square.

So n = 4 gives us 225* 4 = 900 which is a perfect square and 15*4 = 60.

n = 9  gives us 225 * 9 which is a perfect square and 15*9 = 135.

n = 16 gives us 225*16  and 15*16 = 240 , so OK.

n = 25   gives a perfect square and  and 15*25 = 375 - so OK.

n = 36 gives a perfect square and 15*36 = 540 - so OK.

n = 49 gives a perfect square and  15*49 = 735 - so OK.

n = 64 gives a perfect square and 15*64 = 960 - so ok.

n = 81 gives a perfect square and 15 * 81 = 1215  so NOT ok.

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Mathematically, it'll be written as

The null hypothesis is given as

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And the randomization distribution will be centred at 0.1 too.

The alternative hypothesis will now prove the theory they're looking to see in the question that

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Hope this Helps!!!

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Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
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Step-by-step explanation:

The apparent brightness of a star is

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<em>d = distance in ly (light-years) </em>

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Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

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\bf I_1= light intensity at distance \bf d_1  

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Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

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