Answer:
a) 2*4+7*3-10 = 25
b) 3(4-2*3) = -6
c) (5*3-3*4)/12= 4
Step-by-step explanation:
just replace the variables with the numbers given .
Answer:
A: ![v(t)=-t^2+4t-5](https://tex.z-dn.net/?f=v%28t%29%3D-t%5E2%2B4t-5)
Step-by-step explanation:
Acceleration is second derivative of position, velocity is first derivative. Therefore, the velocity is the integral of acceleration.
![v(t)=\int\ {2t+4} \, dt](https://tex.z-dn.net/?f=v%28t%29%3D%5Cint%5C%20%7B2t%2B4%7D%20%5C%2C%20dt)
Integrate:
![-t^2+4t+C](https://tex.z-dn.net/?f=-t%5E2%2B4t%2BC)
V(0)=-5:
![-0^2+4(0)+C=-5\\C=-5](https://tex.z-dn.net/?f=-0%5E2%2B4%280%29%2BC%3D-5%5C%5CC%3D-5)
Therefore, v(t):
![v(t)=-t^2+4t-5](https://tex.z-dn.net/?f=v%28t%29%3D-t%5E2%2B4t-5)
Answer:
99% confidence interval for the soil water content is determined by
(0.19474, 0.26926)
Step-by-step explanation:
<u><em>Explanation</em></u>:-
Given mean of the sample = 23.2% = 0.232
Given standard deviation of the sample = 1.0% =0.01
Given sample size 'n' =4
Degrees of freedom = n-1 =4-1 =3
critical value = 0.01
99% confidence interval for the soil water content is determined by
![((x^{-} - t_{0.005} \frac{S.D}{\sqrt{n} } , (x^{-} + t_{0.005} \frac{S.D}{\sqrt{n} })](https://tex.z-dn.net/?f=%28%28x%5E%7B-%7D%20-%20t_%7B0.005%7D%20%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%2C%20%28x%5E%7B-%7D%20%2B%20t_%7B0.005%7D%20%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%29)
![((0.232 - 7.4534\frac{0.01}{\sqrt{4} } , (0.232 + 7.4534 \frac{0.01}{\sqrt{4} })](https://tex.z-dn.net/?f=%28%280.232%20-%207.4534%5Cfrac%7B0.01%7D%7B%5Csqrt%7B4%7D%20%7D%20%2C%20%280.232%20%2B%207.4534%20%5Cfrac%7B0.01%7D%7B%5Csqrt%7B4%7D%20%7D%29)
( 0.232 -0.03726 , 0.232 +0.03726)
(0.19474, 0.26926)
<u><em>Final answer:</em></u>-
99% confidence interval for the soil water content is determined by
(0.19474, 0.26926)
Answer:
334,65
Step-by-step explanation:
345 / 100 x 97 = 334,65
Answer:
(2,0) x
(0,-6) y
Step-by-step explanation: