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professor190 [17]
4 years ago
8

Solve the inequality please- 21 ≥ 3(a- 7) +9 Thank you so much

Mathematics
1 answer:
Alexeev081 [22]4 years ago
8 0

Answer:

<h2>         a ∈ (-∞, -3></h2>

Step-by-step explanation:

<h3>- 21 ≥ 3(a - 7) + 9</h3><h3>- 21 ≥ 3a - 21 + 9</h3>

  +21         +21

<h3>    0 ≥ 3a + 9 </h3><h3>3a + 9 ≤ 0</h3>

      -9      -9

<h3>3a ≤ - 9</h3>

 ÷3    ÷3

<h3>  a ≤ -3 </h3><h3>a ∈ (-∞, -3></h3>
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Answer:

Probability that this whole shipment will be accepted is 0.7324.

Step-by-step explanation:

We are given that a company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives.

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 = 30 components

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

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<em>LET X = Number of defective components</em>

So, it means X ~ Binom(n=30, p=0.06)

Now, Probability that this whole shipment will be accepted only when there are fewer than 3 defectives = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= \binom{30}{0}\times 0.06^{0} \times (1-0.06)^{30-0}+ \binom{30}{1}\times 0.06^{1} \times (1-0.06)^{30-1}+ \binom{30}{2}\times 0.06^{2} \times (1-0.06)^{30-2}

= 1 \times 1 \times 0.94^{30}+ 30 \times 0.06^{1}  \times 0.94^{29}+ 435 \times 0.06^{2} \times 0.94^{28}

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Therefore, probability that this whole shipment will be accepted is 0.7324.

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I rounded up but the question may call for something different. Hope this Helps!

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