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shutvik [7]
3 years ago
8

Anybody know this ???

Mathematics
1 answer:
serg [7]3 years ago
6 0

Answer:

suurreeeeee let me finish and iwill give you the answer

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Solve the equation for y.
Alex17521 [72]

Answer:

y = \frac{fa - 24}{15}

Step-by-step explanation:

\frac{15y + 24}{a} = f

multiply both sides by a

(15y + 24) = fa

the parenthesis can be removed

15y + 24 = fa

subtract 24 from both sides

15y = fa - 24

divide both sides by 15

y = \frac{fa - 24}{15}

6 0
3 years ago
Write the number in two forms 8.517
Korolek [52]
Below are 8.517 in three forms:

8.517 in mixed number form would be...

8 517/1000

8.517 in fraction form would be...

8517/1000

8.517 in written form would be

Eight and five hundred seventeen thousandths
6 0
3 years ago
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Earth’s diameter is approximately 12,760,000 meters.
Kazeer [188]

Answer:

1.276 × 10^7

Step-by-step explanation:

Just move the decimal until you reach the first number and then count how many times you moved the decimal point. Hope that helps!

5 0
3 years ago
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(i need a fast and clear answer please and ill also mark you as brainlyest)
marishachu [46]

Answer:

Step-by-step explanation:

6) a) Cost = 12.40 x 0.6 = $7.44

   b) Cost = 12.40 x 0.05 = $0.62

8 0
3 years ago
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The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

                                       = 100 × 0.03

                                       = 3

Therefore the we find the probability that there are 3 or less errors on the page as

     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

7 0
3 years ago
Read 2 more answers
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