Answer:
6359
Step-by-step explanation:
add
Answer:
k= -18
Step-by-step explanation:
5 = (k/6) + 8 so 5-8 = k/6
-3=k/6 then -3 multiplied by 6 = k which is -18 = k
-18/6= -3
-3+8=5
The answer to the question
Answer:
B is True
A, C. D are false
Step-by-step explanation:
Given :
Sample size, n = 120
Mean diameter, m = 10
Standard deviation, s = 0.24
Confidence level, Zcritical ; Z0.05/2 = Z0.025 = 1.96
The confidence interval represents how the true mean value compares to a set of values around the mean computed from a set of sample drawn from the population.
The population here is N = 10000
To obtain
Confidence interval (C. I) :
Mean ± margin of error
Margin of Error = Zcritical * s/sqrt(n)
Margin of Error = 1.96 * 0.24/sqrt(120)
Confidence interval for the 10,000 ball bearing :
10 ± 1.96 * (0.24) / sqrt(120)
Hence. The confidence interval defined as :
10 ± 1.96 * (0.24) / sqrt(120) is the 95% confidence interval for the mean diameter of the 10,000 bearings in the box.
The given matrix equation is,
.
Multiplying the matrices with the scalars, the given equation becomes,
![\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%269%5C%5C12%266%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dy%264y%5C%5C3y%262y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20%20)
Adding the matrices,
![\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%2By%269%2B4y%5C%5C12%2B3y%266%2B2y%5Cend%7Barray%7D%5Cright%5D%20%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20)
Matrix equality gives,

Solving the equations together,

We can see that the equations are not consistent.
There is no solution.