Answer;
C) The second arc should be centered at C.
Explanation;
Assuming the goal is to construct a line parallel to AB that passes through given point C.
-Draw a line through C and across AB at an angle creating D.
- With the compass width about half of DC, and center D, draw the first arc to cross both lines.
-Using the same compass width , draw the second arc with center C.
-Then set the compass width to the lower arc (the first arc)
- Move the compass to the second arc. Mark off an arc to make point E
-Draw a straight line through C and E
Thus the line CE will be parallel to line AB
Answer:
Mean = 1.42
Variance = 0.58
Step-by-step explanation:
Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.
Also, joint probability function of X and Y are given.
To find:
mean and variance of X
Solution:
From the given joint probability function of X and Y,

Mean of X:

Variance of X:

![var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58](https://tex.z-dn.net/?f=var%28X%29%3DE%5Cleft%20%5B%20X%5E2%20%5Cright%20%5D-%5Cleft%20%28%20E%5Cleft%20%5B%20X%20%5Cright%20%5D%20%5Cright%20%29%5E2%5C%5C%3D%5Cfrac%7B31%7D%7B12%7D-%5Cleft%20%28%20%5Cfrac%7B17%7D%7B12%7D%20%5Cright%20%29%5E2%5C%5C%3D%5Cfrac%7B31%7D%7B12%7D-%5Cfrac%7B289%7D%7B144%7D%5C%5C%3D%5Cfrac%7B372-289%7D%7B144%7D%5C%5C%3D%5Cfrac%7B83%7D%7B144%7D%5C%5C%3D0.58)
Answer:2/16
Step-by-step explanation:
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701