Answer:
71°
Step-by-step explanation:
The temperature had to be at least 71° when the heater quit to be able to lose 4° and still be at least 67° when the heater is repaired.
Question is...how did someone know that the heater had failed?
Answer:
1. Multiples
2. factors
3.multiples
4.factor
5.factor
Step-by-step explanation:
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:
Area of the sector = pi*(6^2)*(48/360)=4.8*pi= 15.079
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = 
= 
= ![2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-3%2B1%7D%20%7D%7B-3%2B1%7D%5D%5E%7Bb%7D_a%20%20%20dx)
{ Because ![\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a](https://tex.z-dn.net/?f=%5Cint_%7Ba%7D%5E%7Bb%7D%20x%5E%7Bn%7D%20dx%20%3D%20%5B%20%5Cfrac%7Bx%5E%7Bn%2B1%7D%20%7D%7Bn%2B1%7D%5D%5E%7Bb%7D_a)
}
= ![2[ \frac{x^{-2} }{-2}]^{b}_a](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-2%7D%20%7D%7B-2%7D%5D%5E%7Bb%7D_a)
= ![\frac{2}{-2} [ x^{-2} ]^{b}_a](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B-2%7D%20%5B%20x%5E%7B-2%7D%20%5D%5E%7Bb%7D_a)
= ![-1 [ b^{-2} - a^{-2} ]](https://tex.z-dn.net/?f=-1%20%5B%20b%5E%7B-2%7D%20-%20a%5E%7B-2%7D%20%20%5D)
= 
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = 
= 
= 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/
- 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = 
= 
= 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= 
+ (1/
- 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ 
= 0.75
⇒ 
= 1 - 0.75 = 0.25
⇒ 
= 
⇒ = 4 ⇒ x = 
Therefore, value of x such that P(X < x) = 0.75 is 2.
