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2 years ago

Five times the sum of a number and 27 is greater than or equal to six times the sum of that

Mathematics

1 answer:

AleksAgata [21]2 years ago

6 0

The complete version of question:

Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.

Answer:

$5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}$

Step-by-step explanation:

As the description of the statement is:

'Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.

Five times the sum of a number and 27 is written as: $5(x + 27)$

greater than or equal is written as: $\geq$

six times the sum of that number and 26' is written as: 6(x + 26)

so lets combine the whole statement:

$5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)$

solving

$5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)$

$5x+135\ge \:6x+156$

$5x+135-135\ge \:6x+156-135$

$5x\ge \:6x+21$

$5x-6x\ge \:6x+21-6x$

$-x\ge \:21$

$\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}$

$\left(-x\right)\left(-1\right)\le \:21\left(-1\right)$

$x\le \:-21$

Therefore,

$5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}$

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Answer:

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Step-by-step explanation:

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Solution to the problem

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Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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So then is appropiate use the normal distribution to find the probabilities for $\bar X$

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