Answer:
The integral
is 0.
Step-by-step explanation:
A parameterization of curve C can be:
X (t) = cost 0 <= t <= pi
Y (t) = sint 0 <= t <= pi
r (t) = costi + sintj
r '(t) = -sinti + costj
![Fds = [-costsin^3t + sintcos^3t] dt](https://tex.z-dn.net/?f=Fds%20%3D%20%5B-costsin%5E3t%20%2B%20sintcos%5E3t%5D%20dt)
The integral
is given by:
![\int _0^{\pi }\left[-costsin^3t + sintcos^3t dt\right]dt](https://tex.z-dn.net/?f=%5Cint%20_0%5E%7B%5Cpi%20%7D%5Cleft%5B-costsin%5E3t%20%2B%20sintcos%5E3t%20dt%5Cright%5Ddt)

This shows all the work to get it right.
The required expression is 100 - 0.2w
Step-by-step explanation:
Given,
Amount of radioactive substance = 100 mg
The substance decreases by 20% each week.
To find the remaining amount of the in w weeks.
Now,
Each week 20% = 0.2 of the substance will be decreased.
So, in w weeks 0.2×w of the the substance will be decreased.
Hence,
The remaining substance = 100 - 0.2w
Answer:
what number is y?
it would be x no?
Step-by-step explanation:
Answer:
The probability that a randomly chosen light bulb will last less than 900 hours is 0.1587.
Step-by-step explanation:
The life span of these light bulbs is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours
Mean = 
Standard deviation =
We are supposed to find the probability that a randomly chosen light bulb will last less than 900 hours.i.e. P(x<900)
So, 

Z=-1
P(x<900)=P(z<-1)=0.1587
Hence the probability that a randomly chosen light bulb will last less than 900 hours is 0.1587.