Answer:
y= -3
Step-by-step explanation:
I'll assume it is a linear line (straight line).
Sense the y-value doesn't change and stays -3, the equation for the line would be y= -3
And just so you know, the slope of the line would be 0.
Answer:
645 m 86 cm
Step-by-step explanation:
Since 49 cm is greater than 35 cm, we must borrow 100 cm from 800 m, obtaining 799 m 135 cm. Subtracting 154 m 49 cm from that, we get:
645 m 86 cm
The cost of parking is an initial cost plus an hourly cost.
The first hour costs $7.
You need a function for the cost of more than 1 hour,
meaning 2, 3, 4, etc. hours.
Each hour after the first hour costs $5.
1 hour: $7
2 hours: $7 + $5 = 7 + 5 * 1 = 12
3 hours: $7 + $5 + $5 = 7 + 5 * 2 = 17
4 hours: $7 + $5 + $5 + $5 = 7 + 5 * 3 = 22
Notice the pattern above in the middle column.
The number of $5 charges you add is one less than the number of hours.
For 2 hours, you only add one $5 charge.
For 3 hours, you add two $5 charges.
Since the number of hours is x, according to the problem, 1 hour less than the number of hours is x - 1.
The fixed charge is the $7 for the first hour.
Each additional hour is $5, so you multiply 1 less than the number of hours,
x - 1, by 5 and add to 7.
C(x) = 7 + 5(x - 1)
This can be left as it is, or it can be simplified as
C(x) = 7 + 5x - 5
C(x) = 5x + 2
Answer: C(x) = 5x + 2
Check:
For 2 hours: C(2) = 5(2) + 2 = 10 + 2 = 12
For 3 hours: C(3) = 5(3) + 2 = 15 + 2 = 17
For 4 hours: C(3) = 5(4) + 2 = 20 + 2 = 22
Notice that the totals for 2, 3, 4 hours here
are the same as the right column in the table above.
Answer: 0.0129
Step-by-step explanation:
Given : Mean : ![\mu=13.4\text{ years}](https://tex.z-dn.net/?f=%5Cmu%3D13.4%5Ctext%7B%20years%7D)
Standard deviation : ![\sigma=1.7\text{ years}](https://tex.z-dn.net/?f=%5Csigma%3D1.7%5Ctext%7B%20years%7D)
Sample size : ![n=40](https://tex.z-dn.net/?f=n%3D40)
Let X be the random variable that represents the age of of the fleets.
Z-score : ![z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
For x = 14
![z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B14-13.4%7D%7B%5Cdfrac%7B1.7%7D%7B%5Csqrt%7B40%7D%7D%7D%5Capprox2.23)
By using standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be
![P(x\geq14)=P(z\geq2.23)=1-P(z](https://tex.z-dn.net/?f=P%28x%5Cgeq14%29%3DP%28z%5Cgeq2.23%29%3D1-P%28z%3C2.23%29%5C%5C%5C%5C%3D1-0.9871262%3D0.0128738%5Capprox0.0129)
Hence, the the probability that the average age of these 40 airplanes is at least 14 years old =0.0129