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aivan3 [116]
2 years ago
8

Two numbers have a sum of 23 and a difference of 5. What is the smaller of the two numbers.

Mathematics
2 answers:
mr Goodwill [35]2 years ago
8 0
Answer: 14, 9

Step-by-Step Explanation:

Let larger no. be ‘x’
Let smaller no. be ‘y’

Sum (x + y) = 23
Difference (x - y) = 5

Therefore,
x + y = 23 => Eq. 1
x - y = 5 => Eq. 2

On Adding Eq. 1 and 2, we get :-
=> 2x + 0 = 28
=> 2x = 28
=> x = 28/2 = 14

Substitute x = 14 in Eq. 1 :-
=> x + y = 23
=> 14 + y = 23
=> y = 23 - 14 = 9

Therefore, x = 14 ; y = 9

Hence,
Larger number = 14
Smaller number = 9
scZoUnD [109]2 years ago
4 0

Answer:

<h2><u><em>9</em></u> </h2>

Step-by-step explanation:

Two numbers have a sum of 23 and a difference of 5. What is the smaller of the two numbers.

x + y = 23

x - y = 5

(23 - 5) : 2 = 9

9 + 5 = 14

check

14 + 9 = 23

14 - 9 = 5

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Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846

If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours.

This is 1 subtracted by the pvalue of Z when X = 8.1. So

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

By the Central Limit Theorem

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

Z = \frac{8.1 - 8.4}{0.2846}

Z = -1.05

Z = -1.05 has a pvalue of 0.1469

1 - 0.1469 = 0.8531

85.31% probability that their mean rebuild time exceeds 8.1 hours.

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