Answer:
in 4/5 fraction of the game they scored two or more runs.
Let's actually find the line of best fit...
m=(nΣyx-ΣyΣx)/(nΣx^2-ΣxΣx)
m=(11*836-130*55)/(11*385-3025)
m=2046/1210
m=93/55
b=(Σy-93Σx/55)/n
b=(55Σy-93Σx)/(55n)
b=(7150-5115)/(55*11)
b=185/55, so the line of best fit is:
y=(93x+185)/55
A) The approximate y-intercept (the value of y when x=0) is 185/55≈3.36.
Which means that those who do not practice at all will win about 3.36 times
B) y(13)=(93x+185)/55
y(13)≈25.34
So after 13 months of practice one would expect to win about 25.34 times.
Answer:
8h + 15 = 39
Step-by-step explanation:
The base fee is 15 so you shouldn't multiply the hours by it.
The question states 8 per hour which means that you should multiply the 8 by hours and add on the base fee which is 15.
Consider a homogeneous machine of four linear equations in five unknowns are all multiples of 1 non-0 solution. Objective is to give an explanation for the gadget have an answer for each viable preference of constants on the proper facets of the equations.
Yes, it's miles true.
Consider the machine as Ax = 0. in which A is 4x5 matrix.
From given dim Nul A=1. Since, the rank theorem states that
The dimensions of the column space and the row space of a mxn matrix A are equal. This not unusual size, the rank of matrix A, additionally equals the number of pivot positions in A and satisfies the equation
rank A+ dim NulA = n
dim NulA =n- rank A
Rank A = 5 - dim Nul A
Rank A = 4
Thus, the measurement of dim Col A = rank A = five
And since Col A is a subspace of R^4, Col A = R^4.
So, every vector b in R^4 also in Col A, and Ax = b, has an answer for all b. Hence, the structures have an answer for every viable preference of constants on the right aspects of the equations.
