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antiseptic1488 [7]
3 years ago
10

Triangle DEF is similar to triangle ABC, and the length of EF is 5cm. What are the lengths of sides DE is 6cm and DF is 8cm cent

imeters?
Mathematics
1 answer:
ra1l [238]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that,

∆DEF is similar to ∆ABC

∆DEF ≈ ∆ABC

Then, using similar triangle properties

DE / AB = DF /AC = EF / BC

Given that,

EF = 5cm, DE = 6cm and DF = 8cm

6 / AB = 8 / AC = 5 / BC

We need one more data, to solve this problem, so the question is not complete, but let assume AC is 16cm

Then, we can find AB and BC

So, taking each part of the equation

6 / AB = 8 / AC

Since we assumed AC = 16cm

6 / AB = 8 / 16 = ½

6×2 = AB

AB = 12cm

Taking the other parts of the equation

8 / AC = 6 / BC

8 / 16 = 5 / BC

½ = 5 / BC

Bc = 2 × 5

BC = 10cm.

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Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Step-by-step explanation:

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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:

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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.

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In this problem, we have that:

n = 153, p = 0.62

So

\mu = E(X) = np = 153*0.62 = 94.86

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

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

Z = \frac{97 - 94.86}{6}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

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0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

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