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kicyunya [14]
3 years ago
13

Does someone know this answer. My teacher has a hard time solving it .

Mathematics
2 answers:
sukhopar [10]3 years ago
7 0

\sqrt{289}= \\ \sqrt{17\times17}= \\ \sqrt{{17}^{2}}=17

ahrayia [7]3 years ago
4 0

SOS

Answer: \sqrt{289}=17

\mathrm{Factor\:the\:number:\:}\:289=17^2\\=\sqrt{17^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\=17

<em>Hope this helps!!!</em>

<em>Sofia</em>

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Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

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The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 7 parts

            r = number of success = fewer than 2

           p = probability of success which in our question is % of defective

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<em>LET X = Number of parts that are defective</em>

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Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

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                  =  \binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}

                  =  1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}

                  =  <u>0.604</u>

<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>

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