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2 years ago

9. If AJKL - ANML, find the values of x and y.

Mathematics

1 answer:

sergeinik [125]2 years ago

8 0

Answer:

x = 35, y = 18

Step-by-step explanation:

Since both triangles are similar, their sides must be proportional with means all of them have a common ratio.

If NM = 16, and KJ = 40. 40/16 is the ratio from ∆NML to ∆JKL. The reciprocal(numerator and denominator flip, the product of a number and it's reciprocal is 1) of this is true from ∆JKL to ∆NML.

Now to find the missing sides you multiply a side by the common ratio to find it's corresponding side length.

Dividing by a number is the same as multiplying by the reciprocal of that number.

40/16 can be reduced to 5/2 or 2 ½ as a mixed fraction.

16/40 can be reduced to 2/5 or 0.4 as a decimal.

Since the scale of ∆JKL is greater than ∆NML, you multiply by 5/2 to get the corresponding side of ∆JKL from ∆NML.

Since the scale of ∆NML is less than ∆JKL, you multiply by 2/5 to get the corresponding side of ∆NNL from ∆JKL.

Therefore x = 14 × 5/2 = 70/2 = 35.

y = 45 × 2/5 = 90/5 = 18.

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Step-by-step explanation:

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Therefore, the mathematical setting of this question becomes,

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Multiply everything in the bracket with the outside value

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The mathematical setting then becomes,

y+5 = -1/4x + -2 (Note: plus + minus = minus)

so, it becomes

y+5 = -1/4x - 2

The equation is supposed to look like this, y= mx + b

Therefore we add +5 to both sides to cancel out

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1 year ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard

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Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Step-by-step explanation:

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that $\mu = 33208, \sigma = 2503$ .

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when $X = 33208+633 = 33841$ subtracted by the pvalue of Z when $X = 33208 - 633 = 32575$

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57$

X = 33841

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$Z = \frac{33841 - 33208}{357.57}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

X = 32575

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

$Z = \frac{32575- 33208}{357.57}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

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