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77julia77 [94]
3 years ago
6

What is the unit's digit of 3127 to the power of 173

Mathematics
1 answer:
bulgar [2K]3 years ago
8 0
3127^173 ≡ 7^173 mod 10

Now - let's look at the powers of 7 mod 10:

7^1 = 7 mod 10
7^2 = 49 ≡ 9 mod 10
7^3 ≡ 3 mod 10
7^4 ≡ 1 mod 10

And this cycle repeats. So

7^173 ≡ 7^1 ≡ 7 mod 10

The last digit is 7.
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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
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Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

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The standard deviation of the binomial distribution is:

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Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

So

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\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

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