Answer:
1) y = $10 + $2x
2) independent variable is the number of rides (x)
3) dependent variable in this equation is the total cost (y)
Step-by-step explanation:
In this scenario, we have two fixed values and two variables which would be the number of rides (x) and the total cost (y). Therefore, since the entrance fee is a fixed value it is added to the cost of each ride which is multiplied by the number of rides represented by the variable (x).
y = $10 + $2x
In this equation, the independent variable is the number of rides (x) because it is a separate value from the equation and does not depend on any other value. The dependent variable in this equation is the total cost (y) which depends on the number of rides that the individual goes on in order to have its own value.
Answer:
35x^2 - 112x - 364
it's probably wrong since I've only expanded the brackets
Express train travel time:
72x = 18
x = 18/72 = 1/4, or 0.25 hours. A quarter of an hour is 15 minutes
x = 15 minutes
Local Train Travel time:
54x = 18
x = 18/54 = 1/3 hour, or 20 minutes
Adding the stops: 1.5 * 6 = 9 additional minutes
Total Travel time: 20 + 9 = 29
Local train takes 29 - 15 = 14 minutes longer.
Answer is 14 minutes
Answer:
1.42
Step-by-step explanation:
x = original price
x + .06x = 1.50
1.06x = 1.50
x = 1.42
Answer:
A sample of 3851 is required.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.98}{2} = 0.01](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.98%7D%7B2%7D%20%3D%200.01)
Now, we have to find z in the Z-table as such z has a p-value of .
That is z with a pvalue of , so Z = 2.327.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
Variance is 5.76 kWh
This means that ![\sigma = \sqrt{5.76} = 2.4](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B5.76%7D%20%3D%202.4)
They would like the estimate to have a maximum error of 0.09 kWh. How large of a sample is required to estimate the mean usage of electricity?
This is n for which M = 0.09. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.09 = 2.327\frac{2.4}{\sqrt{n}}](https://tex.z-dn.net/?f=0.09%20%3D%202.327%5Cfrac%7B2.4%7D%7B%5Csqrt%7Bn%7D%7D)
![0.09\sqrt{n} = 2.327*2.4](https://tex.z-dn.net/?f=0.09%5Csqrt%7Bn%7D%20%3D%202.327%2A2.4)
![\sqrt{n} = \frac{2.327*2.4}{0.09}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.327%2A2.4%7D%7B0.09%7D)
![(\sqrt{n})^2 = (\frac{2.327*2.4}{0.09})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.327%2A2.4%7D%7B0.09%7D%29%5E2)
![n = 3850.6](https://tex.z-dn.net/?f=n%20%3D%203850.6)
Rounding up:
A sample of 3851 is required.