What numbers can give me 8 when added and 30 when multiplied

1 answer:

7 0

In this problem, it is important to take note that the number of numbers to be utilized isn't specified so it can be up to a thousand numbers. It wasn't also specified if repeating of numbers is allowed or not. So with those taken into consideration and the condition presented in mind, the numbers that can give you 8 when added and 30 when multiplied are 2, 3, 5, -1, and another -1. The derivation from this is mainly from factorization and a little bit of logic.

here is the solution.

2 x 3 x 5 x -1 x -1 = 30
6 x 5 x -1 x -1 = 30
30 x -1 x -1 = 30
-30 x -1 = 30
30 = 30

2 + 3 + 5 + -1 + -1 = 8
5 + 5 + -1 + -1 = 8
10 + -1 + -1 = 8
9 + -1 = 8
8 = 8

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A catering company provides packages for weddings and for showers. The cost per person for small groups

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Using the <em>normal distribution and the central limit theorem</em>, it is found that the probability the mean cost of the weddings is more than the mean cost of the showers is of 0.9665.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

When two variables are subtracted, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances.

<h3>What is the mean and the standard error of the distribution of differences?</h3>

For each sample, they are given by:

$\mu_W = 82.3, s_W = \frac{18.2}{\sqrt{9}} = 6.0667$

$\mu_S = 65, s_S = \frac{17.73}{\sqrt{6}} = 7.2382$

For the distribution of differences, we have that:

$\mu = \mu_W - \mu_S = 82.3 - 65 = 17.3$

$s = \sqrt{s_W^2 + s_S^2} = \sqrt{6.0667^2 + 7.2382^2} = 9.4444$

The probability the mean cost of the weddings is more than the mean cost of the showers is P(X > 0), that is, <u>one subtracted by the p-value of Z when X = 0</u>, hence:

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

By the Central Limit Theorem

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