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evablogger [386]
3 years ago
5

NEED HELP NOW! SOLVE THE INEQUALITY: -r/3 ≤ 6

Mathematics
1 answer:
larisa [96]3 years ago
5 0

Answer:

r\geq -18

Step-by-step explanation:

\frac{-r}{3} \leq 6\\\rule{150}{0.5}\\(\frac{-r}{3})*3\leq 6*3\\\\-r \leq 18\\\\\frac{-r \leq 18}{-1}\\\\\boxed{ r\geq -18}

Hope this helps!

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manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin
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Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) nCr q^{r} p^{n-r}

nCr = \frac{n!}{(n-r)!r!}

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = 20C18 * 0.97^{18} * 0.03^{20-18}

P(X=18) = 20C18 * 0.97^{18} * 0.03^{2}

P(X=18) = 0.0988

P(X=19) = 20C19 * 0.97^{19} * 0.03^{20-19}

P(X=19) = 20C19 * 0.97^{19} * 0.03^{1}

P(X=19) = 0.3364

P(X=20) = 20C20 * 0.97^{20} * 0.03^{20-20}

P(X=20) = 20C20 * 0.97^{20} * 0.03^{0}

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

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3 years ago
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Answer:

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320,040 in expanded form and with names
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300,000+20,000+40

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