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Novay_Z [31]
3 years ago
7

Determine the center of the circle from the equation.

Mathematics
1 answer:
allsm [11]3 years ago
7 0

Answer:

A. (2, -3).

Step-by-step explanation:

The first step is to complete the square on the terms in x and the terms in y:

x² - 4x +4 +y² +6y + 9 =4

= (x - 2)² - 4 + (y + 3)² - 9 + 4 + 9 = 4

= = (x - 2)² + (y + 3)² - 9 + 9 - 4 + 4 = 4

=  (x - 2)² - (y + 3)²  = 4.

So the centre is at (2, -3).

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LOOK AT PICTURE. VOLUME OF CAN PROBLEM
adoni [48]

Answer:

The correct answer is third option.  994

Step-by-step explanation:

<u>Points to remember</u>

Volume of cylinder = πr²h

Where 'r' is the radius and 'h' is the height of cylinder

From the given question we get the cylinder height and radius

The height h = 3 times the diameter of one ball

 = 3 * 7.5 = 22.5 cm

Radius = half of the diameter of a ball

 = 7.5/2 = 3.75 cm

<u>To find the volume of cylinder</u>

Volume of cylinder = πr²h

 = 3.14 * 3.75² * 22.5

 = 993.515 ≈ 994

The correct answer is third option.  994

8 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
Analyzing Patterns This table shows the ratio of parts of red paint to parts of Complete this statement about the ratios. white
Andrew [12]

Answer:

the answer is 3

Step-by-step explanation:

3 0
2 years ago
6. Find both solutions by factoring <br> ): 3x^2 + 10x + 7 = 0
timama [110]
Both solutions:
x = -1, -7/3
7 0
2 years ago
HELP FAST!<br> A) 180 = 30 + (x + 60)<br><br> B) 180 + 30 = x + 60<br><br> C) 90 = 30 + (x + 60)
Fantom [35]
A) 180 = 30 (x+60) :))) of
6 0
3 years ago
Read 2 more answers
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