The smallest value squared would be sqrt0 or 0. Please mark Brainliest!!!
Center : Mean Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5 After the introduction of the new course, center = average(121,134,106,93,149,130,119,128,45) = 113.9 The center has moved to the left (if plotted in a graph) because of the low intake for the new course. Spread before introduction of the new course : Arrange the numbers in ascending order: (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132 Spread = Interquartile range = Q3-Q1 = 19.5 After addition of the new course,
(45,93, 106,119,) 121, (128, 130,134, 149)
Q1=median(45,93,106,119)=99.5
Q3=median (128, 130,134, 149)= 132
Spread = Interquartile range = 132-99.5 =32.5
We see that the spread has increased after the addition of the new course.
Answer:
A) Non-collinear- does not lies on straight line
B) Collinear- lie on straight line
Step-by-step explanation:
We have to check collinearity of three points.
The points
are said to be collinear if,
![\left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] = 0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%26y_1%26z_1%5C%5Cx_2%26y_2%26z_2%5C%5Cx_3%26y_3%26z_3%5Cend%7Barray%7D%5Cright%5D%20%3D%200)
A) A(2, 5, 3), B(3, 7, 1), C(1, 4, 4)
![\left[\begin{array}{ccc}2&5&3\\3&7&1\\1&4&4\end{array}\right] \\\\=2(28-4)-5(12-1)+3(12-7)\\= 8 \neq 0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%265%263%5C%5C3%267%261%5C%5C1%264%264%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D2%2828-4%29-5%2812-1%29%2B3%2812-7%29%5C%5C%3D%208%20%5Cneq%200)
Thus, the given points are not collinear.
B) D(0,-5,5), E(1,-2,4), F(3,4,2)
![\left[\begin{array}{ccc}0&-5&5\\1&-2&4\\3&4&2\end{array}\right] \\\\=0(-4-16)+5(2-12)+5(4+6)\\=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-5%265%5C%5C1%26-2%264%5C%5C3%264%262%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D0%28-4-16%29%2B5%282-12%29%2B5%284%2B6%29%5C%5C%3D0)
Thus, the given points are collinear.
Answer:



Step-by-step explanation:
Given



(0, 1, 0)
The vector equation is given as:

Substitute values for x, y and z

Differentiate:

The parametric value that corresponds to (0, 1, 0) is:

Substitute 0 for t in r'(t)




The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)
It should be noted that:
The equation of a line through position vector a and parallel to vector v is given as:

Such that:
and 
The equation becomes:




By comparison:
and 
The parametric equations for the tangent line are:



I think the answer is 180m