Question:
1. The females worked less than the males, and the female median is close to Q1.
2. There is a high data value that causes the data set to be asymmetrical for the males.
3. There are significant outliers at the high ends of both the males and the females.
4. Both graphs have the required quartiles.
Answer:
The correct option is;
1. The females worked less than the males, and the female median is close to Q1
Step-by-step explanation:
Based on the given data, we have;
For males
Minimum = 0
Q1 = 1
Median or Q2 = 20
Q3 = 25
Maximum = 50
For females;
Minimum = 0
Q1 = 5
Median or Q2 = 6
Q3 = 10
Maximum = 18
Therefore, the values of data that affect the statistical measures of spread and center are that
The females worked less than the males as such the statistical data for the females have less variability than the males in terms of interquartile range
Also the female median is very close to Q1, therefore it affects the definition of a measure of center.
Answer:
See the attached figure which represents the problem.
As shown, AA₁ and BB₁ are the altitudes in acute △ABC.
△AA₁C is a right triangle at A₁
So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)
△BB₁C is a right triangle at B₁
So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)
From (1) and (2)
∴ A₁C/AC = B₁C/BC
using scissors method
∴ A₁C · BC = B₁C · AC
Answer:
y=5x+4
Step-by-step explanation:
First get slope:
(4-9)/(0-1) = -5/-1 = 5
The y-intercept is where the graph hits the y-axis. That is also the point on the graph where x = 0. We are already given that, with point (0,4).
The y-intercept is 4.
y = mx +b, m is slope and b is the intercept.
y = 5x +4
The correct answer is 16/81. Hope this helps.
Answer: √51
—————————
a^2 + b^2 = c^2
a^2 + 7^2 = 10^2
a^2 + 49 = 100
a^2 = 51
√(a^2) = √51
a = √51
