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schepotkina [342]
2 years ago
9

Write a linear function f with f(0)=7 and f(3)=1.

Mathematics
1 answer:
choli [55]2 years ago
7 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

a = ( 0 , 7 ) & b = ( 3 , 1 )

First need to find the slope using the following equation :

slope =  \frac{y(b) - y(a)}{x(b) - x(a)}  \\

slope =  \frac{1 - 7}{3 - 0}  \\

slope =  \frac{ - 6}{3}  \\

slope =  - 2

_________________________________

We have following equation to find the point-slope form of the linear functions using the slope and one of the through points .

y - y(0) = m \times ( \: x - x(0) \: )

x(0) = x - coordinate \:  \: of  \:  \: through \:  \: point \\

y(0) = y - coordinate \:  \: of \:  \: through \:  \: point \\

m = slope

I choose point (( a )) to put in the equation.

y - 7 =  - 2 \times (x - 0)

y - 7 =  - 2x  + 0

y - 7 =  - 2x

Add sides 7

y - 7 + 7 =  - 2x + 7

y =  - 2x + 7

This the slope-intercept form .

Done...

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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
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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.

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A​ 12-sided die is rolled. The set of equally likely outcomes is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} . Find the probability
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Answer:

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Answer: The correct answer is 44.60

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