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3 years ago

|c- 24| = 7c :D :D :D

2 answers:

8 0

Start by making this absolute value equation into two equations; one positive and one negative equation. The two equations will be: c - 24 = 7c and c - 24 = -7c.

Start by solving the positive equation first: c - 24 = 7c. Add 24 to both sides and subtract 7c from both sides of the equation.

-6c = 24, now divide both sides by -6 to find your first c value.

c = -4

Solve the negative equation next: c - 24 = -7c. Add 24 to both sides and add 7c to both sides of the equation.

8c = 24, divide both sides by 8 to isolate c and find your second c value.

c = 3

Substitute to see if these values actually work with the given absolute value equation. Substitute -4 for c.

|c - 24| = 7c ==> |(-4) - 24| = 7(-4)

Solving this equation gives us 28 = -28, and this is a false statement so -4 cannot be part of the solutions. Now check 3 by substituting it for c.

|c - 24| = 7c ==> |(3) - 24| = 7(3)

Solving this equation we get 21 = 21, and this is a true statement so our only possible solution is c = 3.

Ivanshal [37]3 years ago

3 0

$c\geq0\\\\\\ |c- 24| = 7c\\\\c-24=7c \vee c-24=-7c\\\\6c=-24 \vee -8c=-24\\\\c=-4 \vee c=3\\\\-4\not \geq0\\\\\boxed{c=3}$

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a) $P(140$

$P(0.036$

b) $P(140< \bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(139,28.1)$

Where $\mu=139$ and $\sigma=28.1$

We are interested on this probability

$P(140$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(140$

And we can find this probability like this:

$P(0.036$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(0.036$

Part b

For this case we select a sample size of n =32. Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=139, \frac{\sigma}{\sqrt{n}}=\frac{28.1}{\sqrt{32}}=4.97)$

And the new z score would be:

$Z=\frac{\bar X -\mu}{\sigma_{\bar x}}$

$P(140< \bar X$

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