If x = 6, then y = 12. Find y when x = 10.

Car X weighs 136 pounds more than car Z. Car Y weighs 117 pounds more than car Z. The total weight of all three cars is 9439 pou

Aleksandr [31]

Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

$x=z+136$

We also know that Y weighs 117 pounds more than car Z, this can be express as:

$y=z+117$

Finally, we know that the total weight of all the cars is 9439, then we have:

$x+y+z=9439$

Hence, we have the system of the equations:

$\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}$

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

$\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}$

Now that we have the value of z we plug it in the first two equations to find x and y:

$\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}$

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

4 0

1 year ago

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

zaharov [31]

Answer:

a) $P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z$

b) $P(X$

And we can find this probability uing the normal standard table:

$P(z$

c) $P(3.5$

And we can find this probability with this difference:

$P(-1.25$

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

$P(-1.25$

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

$X \sim N(5,1.2)$