The different of anumber and 12

Mathematics

1 answer:

Llana [10]3 years ago

3 0

Five divided by the difference of a number and 12 equals the quotiene of 10 and the sum of the number and five

You might be interested in

How can you write the expression with rationalized denominator? 2+sqrt3(3)/sqrt3(6)

joja [24]

So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root

so

$\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}}
\\\\\\
\cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6}
\\\\\\
\cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}$

8 0

3 years ago

Vail Resorts pays part-time seasonal employees at ski resorts on an hourly basis. At a certain mountain, the hourly rates have a

notsponge [240]

Answer:

$z=0.842$

And if we solve for a we got

$\mu=13.16 +0.842*3=15.69$

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".

Solution to the problem

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

$X \sim N(\mu,3)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>13.16)=0.20$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.20

If we use condition (b) from previous we have this:

$P(X$