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Anit [1.1K]
3 years ago
14

How many pairs (a, b), where a and b are positive integers, satisfy the equation a2−b2=105? PLS HELP

Mathematics
1 answer:
Vlad [161]3 years ago
6 0

Answer:

  4

Step-by-step explanation:

The left side of the equation can be factored as the difference of squares:

  (a -b)(a +b) = 105

The right side of the equation can be factored to 4 different pairs of factors:

  105 = 1·105 = 3·35 = 5·21 = 7·15

For each of the factor pairs, we can match factors to get, for example, ...

  a -b = 1

  a +b = 105

The solution to this is a = (105 +1)/2 = 53, b = (105 -1)/2 = 52. Thus, we have ...

  (a, b) = (53, 52)

For the other factor pairs, the solutions are ...

  (a, b) = (19, 16), (13, 8), (11, 4)

There are a total of 4 positive integer solutions.

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A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of
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Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 15.4 inches, and standard deviation of 3.5 inches.

This means that \mu = 15.4, \sigma = 3.5

16 items are chosen at random

This means that n = 16, s = \frac{3.5}{\sqrt{16}} = 0.875

What is the probability that their mean length is less than 16.8 inches?

This is the p-value of Z when X = 16.8. So

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

By the Central Limit Theorem

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

Z = \frac{16.8 - 15.4}{0.875}

Z = 1.6

Z = 1.6 has a p-value of 0.9452.

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

5 0
3 years ago
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