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.
Given:
The bases of trapezoid measuring 4 m and 12 m.
To find:
The median of the trapezoid.
Solution:
The median of the trapezoid is the average of its bases.
The bases of trapezoid measuring 4 m and 12 m. So, the median of the trapezoid is:
Therefore, the correct option is C.
That number would be 6.
Since 5^2 = 25 and 6^2 = 36, we can see that the only possible whole number between sqrt(27) and sqrt(39) has to be 6.
Answer:
what the question
?
Step-by-step explanation:
Hey there! :)
Answer:
Third option. x = 2, and x = 4.
Step-by-step explanation:
Find the zeros of this quadratic equation by factoring:
f(x) = x² - 6x + 8
Becomes:
f(x) = (x - 4)(x - 2)
Set each factor equal to 0 to solve for the roots;
x - 4 = 0
x = 4
x - 2 = 0
x = 2
Therefore, the zeros of this equation are at x = 2, and x = 4.
