Answer:
1) Event A = 2/3
Event B = 1/2
2) 1/2
Step-by-step explanation:
1)
Event A :
No. we need on dice = 4
Total numbers on dice = 6
Hence sample space of the event = 6
P( getting 4) = 4/6 = 2/3
Event B :
A coin has a head & a tail.
Hence sample space of the event = 2
But as we need tail only ,
P ( getting Tail ) = 1/2 [ if only tossed once ]
2)
Total numbers on a die = 6
Total no. of odd numbers on die = 3 (∵ 1 , 3 & 5 are odd )
Sample space of this event = 6
P (getting an odd number) = 3/6 = 1/2
Amounttotal=amount per day times number of days
200=17.1 times d
divide both sides by 17.1
11.695=d
about 12 days
Well it depends if it's a leap year. If it isn't a leap year there are 28 days and if we are in a leap year than there are 29 days in February.
Hope that helped :)
Answer is a to the question
Answer:
Arc length ![=\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx](https://tex.z-dn.net/?f=%3D%5Cint_0%5E%7B%5Cpi%7D%20%5Csqrt%7B1%2B%5B%284.5sin%284.5x%29%29%5D%5E2%7D%5C%20dx)
Arc length 
Step-by-step explanation:
The arc length of the curve is given by ![\int_a^b \sqrt{1+[f'(x)]^2}\ dx](https://tex.z-dn.net/?f=%5Cint_a%5Eb%20%5Csqrt%7B1%2B%5Bf%27%28x%29%5D%5E2%7D%5C%20dx)
Here, 
interval ![[0, \pi]](https://tex.z-dn.net/?f=%5B0%2C%20%5Cpi%5D)
Now, 
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5Cleft%20%28%20%5B-cos%28t%29%5D_0%5E%7B4.5x%7D%20%5Cright%20%29)
Now, the arc length is ![\int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx](https://tex.z-dn.net/?f=%5Cint_0%5E%7B%5Cpi%7D%20%5Csqrt%7B1%2B%5Bf%27%28x%29%5D%5E2%7D%5C%20dx)
![\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx](https://tex.z-dn.net/?f=%5Cint_0%5E%7B%5Cpi%7D%20%5Csqrt%7B1%2B%5B%284.5sin%284.5x%29%29%5D%5E2%7D%5C%20dx)
After solving, Arc length
