Answer: The area of ABC is 56 m².
Explanation:
It is given that in △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP.
Since point P divides the line AB in 1:3, therefore the area of triangle APC and BPC is also in ratio 1:3. To prove this draw a perpendicular h on AB from C.

Since the area of BPC is
th part of total area, therefore area of APC is
th part of total area.
The point M is the midpoint of CP, therefore the area of BMP and BMC is equal by midpoint theorem.


Area of BPC is,



Area of APC is,



Area of ABC is,


Therefore, the area of ABC is 56 m².
1 litre of 10% solution contains 100mL of iodine
1 litre of 90% solution contains 900mL of iodine
Mixing them will produce 2 litres of solution containing 1000mL of iodine which is 50% iodine.
To produce 8 litres of 50% iodine they can mix 4 litres of 10% and 4 litres of 90%
Answer: 358 square units
Steps:
(12 * 20) + (10 * 10) + (6 * 3)
240 + 100 + 18
358
Out of the 6 numbers, 3 of them are even: 4, 10, and 14. The probability that he will pick an even number is therefore 3/6 = 1/2 = .5
Answer:
The points are randomly scattered with no clear pattern
The number of points is equal to those in the scatterplot.
Step-by-step explanation:
The points in the residual plot of the line of best fit that is a good model for a scatterplot are randomly scattered with no clear pattern (like a line or a curve).
The number of points in the residual plot is always equal to those in the scatterplot.
It doesn't matter if there are about the same number of points above the x-axis as below it, in the residual plot.
The y-coordinates of the points are not the same as the points in the scatterplot.