Answer:
Step-by-step explanation:
5 * 8y = 40y
8 * 5 = 40
40y - 40 IS equal to 40y - 40
Answer:
answer below
Step-by-step explanation:
from the question the
mean = ∑x/n
∑x = 317
n = 16
mean = 19.8125
x x-Ц x-Ц²
19 -0.8125 0.66015625
20 0.1875 0.03515625
19 -0.8125 0.66015625
20 0.1875 0.03515625
18 -1.8125 3.28515625
21 1.1875 1.41015625
17 -2.8125 7.91015625
19 -0.8125 0.66015625
24 4.1875 17.53515625
23 3.1875 10.16015625
21 1.1875 1.41015625
21 1.1875 1.41015625
17 -2.8125 7.91015625
20 -0.8125 0.66015625
20 -0.8125 0.66015625
18 -1.8125 3.28515625
∑(x-Ц)² = 56.44
s.d = 1.9
b.
19.81 + 2*1.9= 23.61
19.81-2*1.9 = 16.01
=
Step-by-step explanation:
A weather balloon is released and rises vertically such that its distance s (t) above the ground during the first 10 sec of flight is given by :
...(1)
(a) We need to find velocity at t = 4 seconds
Differentiate equation (1) wrt t.
![v=\dfrac{ds}{dt}\\\\v=\dfrac{d(6+2t+t^2)}{dt}\\\\v=2+2t](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bds%7D%7Bdt%7D%5C%5C%5C%5Cv%3D%5Cdfrac%7Bd%286%2B2t%2Bt%5E2%29%7D%7Bdt%7D%5C%5C%5C%5Cv%3D2%2B2t)
At t = 4 seconds,
![v=2+2(4)\\\\=10\ ft/s](https://tex.z-dn.net/?f=v%3D2%2B2%284%29%5C%5C%5C%5C%3D10%5C%20ft%2Fs)
At t = 4 seconds, the velocity is 10 ft/s.
(b) When the balloon is 50 feet above the ground,
![6+2t+t^2=50](https://tex.z-dn.net/?f=6%2B2t%2Bt%5E2%3D50)
We need to find t.
![6+2t+t^2=50\\\\t^2+2t-50+6=0\\\\t^2+2t-44=0](https://tex.z-dn.net/?f=6%2B2t%2Bt%5E2%3D50%5C%5C%5C%5Ct%5E2%2B2t-50%2B6%3D0%5C%5C%5C%5Ct%5E2%2B2t-44%3D0)
It is a quadratic equation,
t = 5.708 and t = -7.708
Neglecting negative time, the instant time is 5.708 seconds.
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