Answer:
Logan played for 22 minutes during the second half.
Step-by-step explanation:
Since Keith played the first 22 minutes of a soccer game and Logan then replaced him for the rest of the half, and Logan started the second half and was replaced by Wilson with 18 minutes left in the game, if each half is 40 minutes long To determine how long did Logan play during the second half, the following calculation must be performed:
Second Half Total - Time Played by Wilson = Time Played by Logan
40 - 18 = X
22 = X
Therefore, Logan played for 22 minutes during the second half.
Answer:
The solution to this question can be defined as follows:
Step-by-step explanation:
Please find the complete question in the attached file.
![A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D)
now for given values:
![\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%20-%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%3D0%20%5C%5C%5C%5C)
![\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\](https://tex.z-dn.net/?f=%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B-%5Clambda%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%20-0%5D%20-%200%20-%20%5Cfrac%7B1%7D%7B4%7D%5B0-%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B%28%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Clambda%5E2%20%29%5D%20-%20%5Cfrac%7B1%7D%7B4%7D%5B%5Cfrac%7B%5Clambda%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7D%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B8%7D%5Clambda%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%5Clambda%5E2%20-%20%5Cfrac%7B%5Clambda%5E2%7D%7B2%7D%20-%20%5Clambda%5E3%20-%20%5Cfrac%7B%5Clambda%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%3D0%20%5C%5C%5C%5C%5Cto%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%28%5Clambda%20-%5Cfrac%7B1%7D%7B4%7D%29%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%3D0%5C%5C%5C%5C)
In point b:
Its
spectral radius is less than 1 hence matrix is convergent.
In point c:
![\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\](https://tex.z-dn.net/?f=%5Cto%20c%5E%7B%28k%2B1%29%7D%20%3D%20A%20x%5E%7Bk%7D%2BC%20%5C%5C%5C%5C%5Cto%20x%280%29%20%3D%20%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D3%261%262%5Cend%7Barray%7D%5Cright%29%20%20%2C%20c%20%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%29%5C%5C%5C%5C%20%20%5Cto%20x%5E%7B%28k%2B1%29%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D%20x%5Ek%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%5C%5C)
after solving the value the answer is
:
![\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Clim_%7Bk%20%5Cto%20%5Cinfty%7D%20x%5Ek%3Do%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
Answer:
No, the graph is only increasing while the student rides his bike, rides the bus, and walks. It is stays the same while he waits for the bus and when the bus stops to let him off.
Step-by-step explanation:
Kinetic energy is the energy of an object in motion.
The tennis ball would create kinetic energy as it was both traveling up to it's maximum height and as it was falling back down to the ground.
Because of gravity, the kinetic force would be greater as it was falling back to the ground though.
The answer would be B) just before it reaches the ground.
Answer:
The two numbers of rolls are 25.2 and 46.8.
Step-by-step explanation:
The Chebyshev's theorem states that, if X is a r.v. with mean µ and standard deviation σ, then for any positive number k, we have
Here
Then we know that,
.
Here it is given that mean (µ) = 36 and standard deviation (σ) = 5.4.
Compute the two values between which at least 75% of the contestants lie as follows:
Thus, the two numbers of rolls are 25.2 and 46.8.
