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3 years ago

Factor 6y 2 - 48 + 42y. (6y + 8)(y - 6) 6(y + 8)(y - 1) 6(y - 1)(y - 7)

Mathematics

2 answers:

JulijaS [17]3 years ago

8 0

Answer:

6 (y + 8) (y - 1)

Step-by-step explanation:

astraxan [27]3 years ago

3 0

Answer:

6 (y + 8) (y - 1)

Step-by-step explanation:

Factor the following:

6 y^2 + 42 y - 48

Factor 6 out of 6 y^2 + 42 y - 48:

6 (y^2 + 7 y - 8)

The factors of -8 that sum to 7 are 8 and -1. So, y^2 + 7 y - 8 = (y + 8) (y - 1):

Answer: 6 (y + 8) (y - 1)

PS: it's really helpful to pose the question correctly 6 y^2 + 42 y - 48 NOT

6y 2 - 48 + 42y

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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.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X 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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valkas [14]

Answer:

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Step-by-step explanation:

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