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.
Answer:
Luke's angular speed = 188.49 rad/sec
Inga's angular speed = 188.49 rad/sec
Step-by-step explanation:
Data provided in the question:
Distance between the center and Luke = 17 feet
Distance between the Inga and center = 25 feet
Frequency of rotation, f = 30 rpm
Now,
Angular speed, ω is given as = 2πf
thus,
ω = 2π × 30
or
ω = 188.49 rad/s
here the angular speed is independent of radius
Therefore,
Luke's angular speed = 188.49 rad/sec
Inga's angular speed = 188.49 rad/sec
For the first option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. The range is easily affected by the presence of outliers (data points that are away from other data points). Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Thus, that the range of the two data sets are equal does not mean that the data sets have the same variability. Therefore, the first option is not the correct answer.
For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.
For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.
The equation of the line is y=6x+37
Step-by-step explanation:
You can write the equation using the slope intercept form techniques which is y=mx+c where m is the slope of the line and cis they intercept constant.
Given the point (-6,1) and slope m=6then
Δy/Δx =m
