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Troyanec [42]
2 years ago
7

Please answer correctly !!!! Will mark Brianliest !!!!!!!!!!!!!! Show your work !!!!!!!!!!!

Mathematics
1 answer:
GalinKa [24]2 years ago
7 0

Answer: umm

Step-by-step explanation: uhhhh ok their no work

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Here you go. Hope this helps.

4 0
3 years ago
What is the equation of the line that is parallel to the given
slamgirl [31]

Answer:

y=\frac{1}{5}x+\frac{12}{5}

Step-by-step explanation:

<u><em>Complete Question:</em></u> What is the equation of line that is parallel to the line y=\frac{1}{5}+4 and passes through (-2,2).

Let\ the\ equation\ of\ line\ is\ y=mx+c\\\\Since\ it\ is\ parallel\ to\ the\ line\ y=\frac{1}{5}x+4,\ the\ slope\ of\ both\ the\ lines\ will\ be\ same.\\\\Slope\ of\ y=\frac{1}{5}x+4\ is=\frac{1}{5}\\\\m=\frac{1}{5}\\\\Equation: y=\frac{1}{5}x+c\\\\It\ passes\ through\ (-2,2),\ This\ point\ will\ satisfy\ the\ equation.\\\\2=\frac{1}{5}\times (-2)+c\\\\c=2+\frac{2}{5}\\\\c=\frac{12}{5}\\\\The\ Equation\ is: y=\frac{1}{5}x+\frac{12}{5}

7 0
2 years ago
Solve for x<br> 3x+5≡0 mod 9
Whitepunk [10]

3x+5=0

3x=0-5

3x=-5

Divide both sides by 3

-1.6666.......

Hope it helps.

As for mod 9, I don't understand.

5 0
2 years ago
Read 2 more answers
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
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Answer:

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