Answer:
1) a. False, adding a multiple of one column to another does not change the value of the determinant.
2) d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
Step-by-step explanation:
1) If the multiple of one column of a matrix A is added to another to form matrix B then we get: |A| = |B|. Here, the value of the determinant does not change. The correct option is A
a. False, adding a multiple of one column to another does not change the value of the determinant.
2) Two matrices can be column-equivalent when one matrix is changed to the other using a sequence of elementary column operations. Correc option is d.
d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.
Answer:
(5x = 2 = 3 +8a
Step-by-step explanation:
15u + 11u - 9d - 19d = -2d
It went up 26, down 28, a net change of down 2.
Answer:
A. factor 4
Step-by-step explanation:
Just choose any side of the original one e.g 60mm and divide it by the lenght of the same side of the new one. In A. this gives you 60mm/15mm = 4. Then you can multiply the other lengths with that factor and if the result is equal to the sides of your original one you found the solution.
Answer:
(a) 0.06154
(b) 0.2389
(c) 0.6052
(d) 2478
Step-by-step explanation:
probability density function of the time to failure of an electronic component in a copier (in hours) is
P(x) = 1/1076e^−x/1076
λ = 1/1076
A) A component lasts more than 3000 hours before failure:
P(x>3000) = 1 − e^−3000/1076
= 0.06154
B) A component fails in the interval from 1000 to 2000 hours:
P(1000>x>2000) =1 − e^−2000/1076 − 1 +e^−1000/1076 = e^−1000/1076 − e^−2000/1076 = 0.3948 − 0.1559
= 0.2389
C) A component fails before 1000 hours:
P(x<1000) = 0.6052
D) The number of hours at which 10% of all components have failed:
e^−x/1076 = 0.1
= −x/1076
= ln(0.1)
x =(2.3026)×(1076)
x = 2478