Find the inverse of the function f(x) = −5x<br><br> The inverse is g(x) =

Matter is in a liquid state when its temperature is between its melting point and its boiling point. Suppose that some substance

a_sh-v [17]

Answer:

i. When the temperature is above $616.712^{o}$ F, it changes to gas.

ii. When the temperature is below $-33.052^{o}$ F, it changes to solid.

Step-by-step explanation:

Matter generally exists in either a solid, liquid or gaseous form. With each phase having a certain range of temperature.

Temperature scale of a given substance can be either in Celsius, Fahrenheit or Kelvin. And conversion from one scale to another can be achieved. Example, Celsius scale can be converted to Fahrenheit by:

F = $\frac{9}{5}$ θ + 32

where: F is the equivalent temperature in Fahrenheit, θ is the value of temperature in degree Celsius.

Given that: melting point of the substance = $-36.14^{o}$ C

⇒ F = $\frac{9}{5}$ x $-36.14^{o}$ + 32

= $-33.052^{o}$ F

The boiling point = $324.84^{o}$ C

F = $\frac{9}{5}$ x $324.84^{o}$ + 32

= $616.712^{o}$ F

The melting point of the substance is $-33.052^{o}$ F, and boiling point is $616.712^{o}$ F.

Therefore, the range of temperatures for which the substance is not in a liquid state are:

i. When the temperature is above $616.712^{o}$ F, it changes to gas.

ii. When the temperature is below $-33.052^{o}$ F, it changes to solid.

3 0

3 years ago

The distribution of a sample of the outside diameters of PVC pipes approximates a normal distribution. The mean is 14.0 inches,

lakkis [162]

Answer:

A. 13.9 and 14.1 inches

See explanation below.

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 outside diameters of a population, and for this case we know the distribution for X is given by:

$X \sim N(14,0.1)$

Where $\mu=14$ and $\sigma=0.1$

If we want the middle 68% of the data we need to have on the tails 16% on each one

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

$P(X>a)=0.84$ (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 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84

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

$P(X$