Ok, so based on a survey - you know that out of 30 students, 11 said that their favourite subject is mathematics. :-)
This means that:
11/30 of these students think mathematics if their favourite subject.
Now, in order to answer your question, we are going to have to change the denominator of the fraction above to 1000. Whatever we do to the denominator of this fraction, we must do to the numerator of this fraction.
Example:
This means that rounded to the nearest ten, theoretically speaking, 370 students out of 1000 would say that mathematics is their favourite subject.
What is the mean, median, and mode of 6,2,3,4,4,2,8,5,2
pogonyaev
Mean: 4
median: 4
mode: 2
range: 6
You can solve this either just plain algebra or with the use of trigonometry.
In this case, we'll just use algebra.
So, if we let M be the the point that partitions the segment into a ratio of 3:2, we have this relation:
KM/ML = 3/2
KM = 1.5 ML
We also have this:
KL = KM + ML
Substituting KM,
KL = (3/2) ML + ML
KL = 2.5 ML
Using the distance formula and the given coordinates of the K and L, we get the length of KL
KL = sqrt ( (5-(-5)^2 + (1-(-4))^2 ) = 5 sqrt(5)
Since,
KL = 2.5 ML
Substituting KL,
ML = (1/2.5) KL = (1/2.5) 5 sqrt(5) = 2 sqrt(5)
Using again the distance formula from M to L and letting (x,y) as the coordinates of the point M
ML = 2 sqrt(5) = sqrt ( (5-x)^2 + (1-y)^2 ) [let this be equation 1]
In order to solve this, we need to find an expression of y in terms of x. We can use the equation of the line KL.
The slope m is:
m = (1-(-4))/(5-(-5) = 0.5
Using the general form of the linear equation:
y = mx +b
We substitue m and the coordinate of K or L. We'll just use K.
-5 = (0.5)(-4) + b
b = -1.5
So equation of the line is
y = 0.5x - 1.5 [let this be equation 2]
Substitute equation 2 to equation 1 and solving for x, we get 2 values of x,
x=1, x=9
Since 9 does not make sense (it does not lie on the line), we choose x=1.
Using the equation of the line, we get y which is -1.
So, we get the coordinates of point M which is (1,-1)
No there is nothing here please provide answers
2y² + y - 3
(2y+3)(y-1)
2y(y-1) + 3(y-1)
2y² - 2y + 3y - 3
2y² + y - 3
2y + 3 = 0
2y = -3
y = -3/2
y = -1 1/2
y - 1 = 0
y = 1
2y² + y - 3 = 0
2(-3/2)² + (-3/2) - 3 = 0
2(9/4) - 3/2 - 3 = 0
18/4 - 3/2 - 3 = 0
18/4 - (3/2 * 2/2) - 3(4/4) = 0
18/4 - 6/4 - 12/4 = 0
(18 - 6 - 12)/4 = 0
(18 -18)/4 = 0
0/4 = 0
0 = 0
