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IceJOKER [234]
1 year ago
9

a hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. the c

hord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$, where $m$ and $n$ are relatively prime positive integers. find $m n$.
Mathematics
1 answer:
allochka39001 [22]1 year ago
4 0

According to given conditions, m+n is equal to 409.

Consider the diagram below.

In the hexagon ABCDEF, let

AB = BC = CD = 3;

DE = EF = FA = 5;

Arc BAF is equal to one-third of the circle's circumference.

Hence, ∠BCF = ∠BEF = 60°;

Similarly, ∠CBE = ∠CFE = 60°;

Let the point of intersection of BE and CF be P, BE and AD be Q and CF and AD be R.

∴ Δ EFP and Δ BCP are equilateral, and so Δ PQR is also equilateral.

Also, ∠ BAD and ∠ BED subtend the same arc and so do ∠ ABE and ∠ ADE.

∴ Δ ABQ is similar to Δ EDQ

\frac{AQ}{EQ}  = \frac{BQ}{DQ} = \frac{AB}{ED} = \frac{3}{5}

Also,

\frac{\frac{AD - PQ}{2} }{PQ + 5} = \frac{3}{5}  

and \frac{3 - PQ}{\frac{AD + PQ}{2} }  = \frac{3}{5}

On solving these simultaneous equations, we get AD = 360/49

∴ m + n = 409.

To learn more about similarity of triangles, refer to this link:

brainly.com/question/25882965

#SPJ4

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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

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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

\mu = E(X) = np = 110*0.2 = 22

\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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