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storchak [24]
3 years ago
14

Which equation represents a circle that contains the point (-5,-3) and has a center at (-2,1)

Mathematics
1 answer:
MatroZZZ [7]3 years ago
4 0

Answer:

The equation of circle is:

Solution:

The general form of equation is given as:

Where, "r" is radius of circle and (a, b) is center of circle

Given that center at (-2, 1)

a = -2 and b = 1

Substitute (a , b) = (-2, 1) and (x, y) = (-5, -3) in general equation

Substitute r = 5 and (a, b) = (-2, 1) in general equation

Thus the equation of circle is found

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Determine the XNY intercepts of the graph of X+2y=-4
podryga [215]

Answer:

x intercepts (-4,0)

y intercepts (0, -2)

4 0
3 years ago
Classify the real numbers choose all that apply
denis-greek [22]
Rational because 5^3



irrational
4 0
3 years ago
Two consecutive odd integers have a sum of 44. Find the integers.
ivanzaharov [21]

Step-by-step explanation:

Answer

Let2odintegers

(2n+1),(2n+3)

→(2n+1)+(2n+3)=44

⇒4n+4=44

⇒4n=40

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then,integerare

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7 0
2 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
PLEASE HELP ASAP !!!!!!!!
AnnyKZ [126]
Answer: 7°

Explanation:

FBD = EBC
2x + 3 = 9x - 11
-7x = -14
x = -14/-7
x = 2

EBC = 9(2) - 11 = 18 - 11 = 7°
7 0
3 years ago
Read 2 more answers
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