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fgiga [73]
2 years ago
10

Determine the domain of the function, and then graph it.

Mathematics
1 answer:
Tomtit [17]2 years ago
6 0
First set change the function f(x) to y so
that it would be
y = \sqrt{x + 4} - 1
Then set y = 0
0 = \sqrt{x + 4} - 1
Then solve for x

X = - 3

To graph it, just plot the point (-3,0) on the x-axis
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Decrease $250 by 34%
BARSIC [14]

New amount = starting value x (1- percent)

New amount = 250 x (1-0.34)

New amount = 250 x 0.66

New amount = 165

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There are 3,785 milliliters in 1 gallon, and there are 4 quarts in 1 gallon. How many milliliters are in 3 gallons?
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Answer: 11355ml

Step-by-step explanation: It gives you how many milliliters are in one gallon.

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Write a ratio this is equivalent to 9/27
tresset_1 [31]
9/27

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one if the ratio is 1/3
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3 years ago
The point P on the circle x2 + y2 = r2 that is also on the terminal side of an angle θ in standard position is given. Find the i
Brut [27]
Okay so you do not really need the circle equation. if you make a triangle with it's x length as 3 and it's y length as 4, you will be able to find the third length. Do Pythagorean theorem to find the hypotenuse. 
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sin is opposite over hypotenuse or Y over R

so... the sin is 4/5

make sure you know what  quadrant your triangle is in for the negatives
 
6 0
2 years ago
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
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