Given that f(x) = 4x - 3 and g(x) 2x-1 divided by 3<br><br> solve for g(f(2))

Khan academy how to solve equations with fractions and mixed numbers

Rufina [12.5K]

Generally, you are told to approach these by "clearing fractions". That is, you generally multiply the equations by the least common denominator so all fractions and mixed numbers become integers.

Alternatively, you can simply do the arithmetic using the numbers given. You learned a long time ago how to add, subtract, multiply, and divide mixed numbers and fractions. Do these operations as necessary to solve the equations.

8 0

3 years ago

The average THC content of marijuana sold on the street is 9.3%. Suppose the THC content is normally distributed with standard d

velikii [3]

Answer:

a) $X \sim N(9.3,1)$

b) $P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

c) The value of height that separates the bottom 75% of data from the top 25% is 9.9745.

Step-by-step explanation:

1) 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".

2) Part a

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

$X \sim N(9.3,1)$

Where $\mu=9.3$ and $\sigma=1$

3) Part b

We are interested on this probability

$P(X>9.2)$

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:

$P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

4) Part c

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

$P(X>a)=0.25$ (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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.6745. On this case P(Z<0.6745)=0.75 and P(z>0.6745)=0.25

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

$P(X$