![h(t)=-4.9t^{2}+34.3t+1 \\h'(t)=(-4.9t^{2}+34.3t+1)'=-9.8t+34.3 \\h'(t)=0 \\-9.8t+34.3=0 \\ \\t= \frac{34.3}{9.8} =3.5 \\ \\h_{max}=h(3.5)=-4.9\times3.5^{2}+34.3\times3.5+1=61.025](https://tex.z-dn.net/?f=h%28t%29%3D-4.9t%5E%7B2%7D%2B34.3t%2B1%0A%5C%5Ch%27%28t%29%3D%28-4.9t%5E%7B2%7D%2B34.3t%2B1%29%27%3D-9.8t%2B34.3%0A%5C%5Ch%27%28t%29%3D0%0A%5C%5C-9.8t%2B34.3%3D0%0A%5C%5C%0A%5C%5Ct%3D%20%5Cfrac%7B34.3%7D%7B9.8%7D%20%3D3.5%0A%5C%5C%0A%5C%5Ch_%7Bmax%7D%3Dh%283.5%29%3D-4.9%5Ctimes3.5%5E%7B2%7D%2B34.3%5Ctimes3.5%2B1%3D61.025)
The ball reaches its maximum height after 3.5 seconds, and the maximum height is 61.025 meters.
Answer:
Step-by-step explanation:
Multiply 3 times 4x which gives you 12. Multiply 3 times 2 which gives you 6. Your final answer is 12x+6
Answer:
Its A.
Step-by-step explanation:
I just finsihed that question.
Answer: how many hours is she working
Step-by-step explanation:
Answer: a) 0.2222, b) 0.3292, c) 0.1111
Step-by-step explanation:
Since we have given that
Let the probability of getting head be p.
Since, its head is twice as likely to occur as its tail.
![p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}](https://tex.z-dn.net/?f=p%2B%5Cdfrac%7Bp%7D%7B2%7D%3D1%5C%5C%5C%5C%5Cdfrac%7B3p%7D%7B2%7D%3D1%5C%5C%5C%5Cp%3D%5Cdfrac%7B2%7D%7B3%7D)
a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?
So, here, n = 3
![p=\dfrac{2}{3}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B2%7D%7B3%7D)
![q=\dfrac{1}{3}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7B1%7D%7B3%7D)
Now,
![P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5E3C_1%28%5Cdfrac%7B2%7D%7B3%7D%29%5E1%28%5Cdfrac%7B1%7D%7B3%7D%29%5E2%3D0.2222)
b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?
2 tails means 3 heads.
So, it becomes,
![P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5E5C_3%28%5Cdfrac%7B2%7D%7B3%7D%29%5E3%28%5Cdfrac%7B1%7D%7B3%7D%29%5E2%3D0.3292)
c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?
![P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%3D%5Csum%20_%7Bx%3D0%7D%5E1%5E4C_x%28%5Cdfrac%7B2%7D%7B3%7D%5Ex%28%5Cdfrac%7B1%7D%7B3%7D%29%5E%7B4-x%7D%3D0.1111)
Hence, a) 0.2222, b) 0.3292, c) 0.1111