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oksano4ka [1.4K]
4 years ago
14

(-2,3)(0,8) what is en equation for these points

Mathematics
1 answer:
jenyasd209 [6]4 years ago
4 0

Answer:

The line equation that passes through the given points is          5x – 2y + 16 = 0

Explanation:

Given:

Two points are A(-2, 3) and B(0, 8).

To find:

The line equation that passes through the given two points.

Solution:

We know that, general equation of a line passing through two points (x1, y1), (x2, y2) is given by

\frac{(y- y1)}{(x-x_1)}= \frac{((y_2- y_1)}{(x_2- x_1 )}

{(y- y1)= \frac{((y_2- y_1)}{(x_2- x_1 )}\times(x-x_1).............(1)

here, in our problem x1 = 0, y1 = 8, x2 = -2 and y2 = 3.

Now substitute the values in (1)

y-8 = \frac{(3-8)}{(- 2 - 0)}\times(x- 0)

y-8 = \frac{(- 5)}{(-2)}\times(x)

y-8 =\frac{5}{2}x

2y – 16 = 5x

5x – 2y + 16 = 0

Hence, the line equation that passes through the given points is 5x – 2y + 16 = 0.

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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
What are some examples of Adjacent Congruent Angles.
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Answer:

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Step-by-step explanation:

When it comes to angles, the term "adjacent" means different things in different contexts. Any pair of angles in a triangle are considered to be adjacent. The base angles of an isosceles triangle are adjacent congruent angles.

Angles at successive corners of any regular polygon are congruent adjacent angles.

Angle that have a common vertex and a common side are also called adjacent angles. Angles on either side of an angle bisector are adjacent congruent angles. Likewise any non-vertical pair of angles where lines cross at right angles are adjacent congruent angles.

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