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antoniya [11.8K]
3 years ago
6

-3x + 4y = 95 x + 5y = 38

Mathematics
1 answer:
NARA [144]3 years ago
7 0
We can solve this by elimination, in order to do this we need to multiply the bottom equation by three to cross out the X

(x+5y=38)3 3x+15y=114

-3x+4y=95
3x+15y=114

then add the two equations together and the X will cancel out

19y=209
-----  -----
19     19

y=11

then plug that in into one of the equations

x+55=38
  -55   -55

x= -17

your answer is (-17,11)
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If m (AB = 120°, then which of the following have measures of 60°?
victus00 [196]
<span>∠BCA = 120 /2 = 60

answer
</span><span>∠BCA</span>
7 0
2 years ago
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On a coordinate plane, point A is located at (7, 4), and point B is located at (-8, 4). What is the distance between the two poi
wel

Answer:

Solution given;

A(x1,y1)=(7,4)

B(x2,y2)=(-8,4)

distance AB=?

we have

d=\sqrt{(x2-x1) ²+(y2-y1) ²}

AB=\sqrt{(-8-7)²+(4-4)²}

AB=15 units

the distance between the two points is 15 units.

4 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
Helpppppppppppppppppppppplppoooooo
Komok [63]

Answer:

hope it helps you plz mark me brilliantest

Step-by-step explanation:

B

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Dovator [93]

There is some ambiguity here which could be removed by using parentheses.  I'm going to assume that you actually meant:

               x-3

h(x) = ---------------

          (x^3-36x)

To determine the domain of this function, factor the denominator:

x^3 - 36x = x(x^2 - 36) = x(x-6)(x+6)

The given function h(x) is undefined when the denominator = 0, which happens at {-6, 0, 6}.

Thus, the domain is "the set of all real numbers not equal to -6, 0 or 6."

Symbolically, the domain is   (-infinity, -6) ∪ (-6, 0) ∪ (0, 6) ∪ (6, +infinity).

3 0
3 years ago
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