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3 years ago

Please help on this math question

Mathematics

1 answer:

Llana [10]3 years ago

4 0

Answer:

$\large\boxed{C.\ \dfrac{-2x^2-x+72}{8x^2}}$

Step-by-step explanation:

$\dfrac{9}{x^2}-\dfrac{2x+1}{8x}=\dfrac{9\cdot8}{x^2\cdot8}-\dfrac{(2x+1)\cdot x}{8x\cdot x}\qquad\text{use the distributive property}\\\\=\dfrac{72}{8x^2}-\dfrac{2x^2+x}{8x^2}=\dfrac{72-(2x^2+x)}{8x^2}=\dfrac{72-2x^2-x}{8x^2}\\\\=\dfrac{-2x^2-x+72}{8x^2}$

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A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as

Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that $p = 0.17$

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891$

0.3891 = 38.91% probability that only one is a second

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3 years ago

Math need help please

AnnyKZ [126]

Answer:

none

Step-by-step explanation:

4x - 6 = 2(2x + 3)

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