Answer:
The answer is "
f and g are arbitrary".
Step-by-step explanation:
The matrix of the device is increased
![\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%26f%5C%5C%20c%26d%26g%5C%5C%20%5Cend%7Barray%7D%5Cright%5D)
Reduce the echelon row matrix
![\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right] \\\\R_1 \leftrightarrow R_2 \\\\\frac{R_2 -1}{C R_1 \to R_2} \sim \left[\begin{array}{ccc} c&d&g\\ 0 & \frac{3c-d}{c}& \frac{cf-g}{c}\\ \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%26f%5C%5C%20c%26d%26g%5C%5C%20%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CR_1%20%5Cleftrightarrow%20%20R_2%20%5C%5C%5C%5C%5Cfrac%7BR_2%20-1%7D%7BC%20R_1%20%5Cto%20R_2%7D%20%20%20%5Csim%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20c%26d%26g%5C%5C%200%20%26%20%5Cfrac%7B3c-d%7D%7Bc%7D%26%20%5Cfrac%7Bcf-g%7D%7Bc%7D%5C%5C%20%5Cend%7Barray%7D%5Cright%5D)
Therefore, if 3c
0 is d 3c, the device is valid. Therefore d are arbitrary 3c, g and f.
Answer:
Carla has 28 students in her class and Linda has 21
Step-by-step explanation:
Because I just took 49 and substracted a number about halfway and then since carla and 7 more students you just had to find the number that has seven more than linda.
What is radius of the circle and the height of the cylinder.
Answer:
16%
Step-by-step explanation:
The indicated probability is actually the area under the standard normal curve to the left of the mean. I used the function normalcdf( on my TI-83 Plus calculator to find this quantity:
normalcdf(-1000,85,90,5) = 0.1587.
Note #1: This quantity (area / probability) is the area to the left of 85.
Note #2: by the Empirical Rule, 68% of data lies within 1 s. d. of the mean, so the area between the mean (90) and the score (85) is half of 68%, or 34%. Subtracting this from 50% (the area to the left of the mean), we get 16%, which is roughly equivalent to the 0.1587 we got earlier.
Answer:
True
Step-by-step explanation:
