Step-by-step explanation:
Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me.Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me. Essay on I can and I will. Watch me
Answer:
Step-by-step explanation:
R-value can range from -1 to 1 with correlation being strongest at -1 and 1 and weakest at 0.
So the answer is D. -0.9.
Answer:
One number is 17 and the other is 85.
Step-by-step explanation:
Let the numbers be m and n. Let m = (1/5)n. Note that m+n = 18.
Substituting (1/5)n for m in the 2nd equation, we get (1/5)n + (5/5)n = 18. Multiplying all three terms by 5 to remove fractions, we get n + 5 = 90, so that n = 85. m is (1/5) of 85, or m = 17.
Setup
remember is means =, and of means *
Writing across the words
15 = x * 39
15/39 = x
x =0.3846 = 38.46%
Answer:
Only d) is false.
Step-by-step explanation:
Let be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that . 0 is a root of p, so we have that .
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.