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kakasveta [241]
3 years ago
10

The object distance for a concave lens is 2.0 cm, and the image distance is 8.0 cm. The height of the object is 1.0 cm. What is

the height of the image?
1.0 cm
4.0 cm
0.25 cm
16 cm
Mathematics
1 answer:
nika2105 [10]3 years ago
3 0
For the answer to the question, what is the height of the image if t<span>he object distance for a concave lens is 2.0 cm, and the image distance is 8.0 cm. The height of the object is 1.0 cm.

The answer is 4.0cm

I hope my answer helped you. Have a nice day!</span>
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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 260.5-cm and a standard dev
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Answer:

P(M > 260.2-cm) = 0.702

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 260.5, \sigma = 1.6, n = 8, s = \frac{1.6}{\sqrt{8}} = 0.5657

P(M > 260.2-cm)

This is 1 subtracted by the pvalue of Z when X = 260.2. So

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

By the Central Limit Theorem

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

Z = \frac{260.2 - 260.5}{0.5657}

Z = -0.53

Z = -0.53 has a pvalue of 0.298.

1 - 0.298 = 0.702

So

P(M > 260.2-cm) = 0.702

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