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UNO [17]
4 years ago
6

If the scale of a flag from the image to the actual were 1 inch = 63.75 ft, what would the actual width of a large American flag

measure if on the image the flag is 4 inches wide?
Mathematics
1 answer:
velikii [3]4 years ago
6 0

Answer:

21.25 feet or 255 inches.

Step-by-step explanation:

The first step in this problem is to covert each measurement to the same system of measurement (feet, inches, meters, etc.).

Let's convert 4 inches to feet.

As there are 12 inches in a foot, and 4/12 = 1/3, 4 inches = 1/3 foot.

Now, multiply 1/3 foot by 63 3/4 (the amount of feet in mixed number form).

63 3/4 * 1/3 = 21 1/4, or 21.25.

Therefore, the final answer is 21.25 feet.

If you want the answer in inches, all you have to do is multiply 21.25 by 12, as there are 12 inches in a foot.

21.25*12 = 255.

So, the alternative answer is 255 inches.

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Suppose that minor errors occur on a computer in a space station, which will require re-calculation. Assume the occurrence of er
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Answer:

a

  P(X = 0) =  0.6065

b

P(x <  25 ) =   1.18 *10^{-33}

c

 P(x \le 5 ) =  0.9994    

Step-by-step explanation:

From the question we are told that

  The rate  is \lambda =  \frac{1}{2}\   hr^{-1}    =  0.5 / hr

  Generally  Poisson distribution formula is mathematically represented as

       P(X = x) =  \frac{(\lambda t) ^x e^{-\lambda t }}{x!}

Generally the probability that no error occurred during a day is mathematically represented as  

Here  t =  1  hour according to question a

So

   P(X = x) =  \frac{\lambda^x e^{-\lambda}}{x!}

Hence

   [tex]P(X = 0) =  \frac{\frac{1}{2} ^0 e^{-\frac{1}{2}}}{0!}

=>  P(X = 0) =  0.6065

Generally the probability that  a critical error occurs since the start of a day is mathematically represented as

Here  t =  1  hour according to question a

So

   P(X = x) =  \frac{\lambda^x e^{-\lambda}}{x!}

Hence

      P(x \ge 25 ) =  1 - P(x <  25 )

Here

     P(x <  25 ) = \sum_{x=0}^{24} \frac{e^{-\lambda} * \lambda^{x}}{x!}

=>   P(x <  25 ) =  \frac{e^{-0.5} *0.5^{0}}{0!} + \cdots + \frac{e^{-0.5} *0.5^{24}}{24!}

P(x <  25 ) =  0.6065 + \cdots + \frac{e^{-0.5} *0.5^{24}}{6.204484 * 10^{23}}

P(x <  25 ) =  0.6065 + \cdots + 6.0*10^{-32}

P(x <  25 ) =   1.18 *10^{-33}

Considering question c

Here  t =  2  

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 P(x \le 5 ) =  0.9994    

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