Answer:
Step-by-step explanation:
<u>Actual question is:</u>
- logf 4(3d) + log4 1 = log4 3d
<u>Solution:</u>
- logf 4 + logf (3d) + 0 = log4 (3d)
- logf 4 + logf (3d) = log4 (3d)
<u>If we assume f = 4, then</u>
- log4 4 + log4 (3d) = log4 (3d)
- log4 4 = 0
but log4 4 = 1
Therefore the statement is False
2. 5 hours
3. $200
4. 9.5 hours. If it has to be whole numbers, 10 hours
Hope that helps!
<em>-scsb17hm</em>
Hello,
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
![l= 2*5* \int\limits^{\pi}_0{sin( \frac{t}{2} )} \, dt= 5[-cos (t/2)]_0^{\pi}\\\\ =5(0+1)=5](https://tex.z-dn.net/?f=l%3D%202%2A5%2A%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%7Bsin%28%20%5Cfrac%7Bt%7D%7B2%7D%20%29%7D%20%5C%2C%20dt%3D%205%5B-cos%20%28t%2F2%29%5D_0%5E%7B%5Cpi%7D%5C%5C%5C%5C%0A%3D5%280%2B1%29%3D5)
Here is the method but i may have make some mistakes.
Answer:
The number of viewers Network A expects will watch their show is 1.4 million viewers.
Step-by-step explanation:
The expected value is calculated by multiplying the possible outcomes by the probability of their occurrence and adding the results
Therefore, we have the expected value given by the following expression;
Estimated network A viewers where network B schedule top show = 1.1 million viewers
Estimated network A viewers where network B schedule a different show = 1.6 million viewers
Probability that Network B will air its top show = 0.4
Probability that Network B will air another show = 0.6
We therefore have;
Expected value, E of Network A viewers is therefore;
E = 1.1 × 0.4 + 1.6 × 0.6 = 0.44 + 0.96 = 1.4 million viewers.
Network A expects 1.4 million viewers will watch their show.
