Answer:
74/8
Step-by-step explanation:
8 10
8
Multiply The denominator by whole number then add numerator which is 74
then put that number over the original denominator which is 8
Answer:
D) y = -2
The equation of the perpendicular line to the given line is y = -2
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given point (-3,-2)
In graph the equation of line x = 8 ( parallel to y- axis )
1 x + o y - 8 =0
The equation of the perpendicular line to the given line
b x - a y + k =0
o x - 1 y + k =0 ...(i)
The equation (i) is Paases through the point ( -3,-2)
0 ( -3) -1(-2) + k =0
2 + k =0
k = -2
∴<em>The equation of the perpendicular line to the given line is </em>
<em> put k = -2 in equation (i) </em>
o x - 1 y + k =0
- y +(-2) =0
- y = 2
y = -2
<u><em>Final answer:-</em></u>
∴The equation of the perpendicular line to the given line is y =-2
Answer:
Sue's scores for the four games in ascending order are: 97, 98, 98, 107
Step-by-step explanation:
Her modal score was 98. The mode is found by using the number that appears most often. This means that 98 has to appear at least two times out of the four scores.
Her range was 10. The range is found by taking the highest score and subtracting it from the lowest score. The highest score had to be greater than 98 and the lowest score had to be less than 98 since we know the mode was 98.
Her mean score was 100. This mean is found by adding all the numbers together and then dividing by the total numbers listed. Adding the four scores together and dividing by 4 will equal 100.
Used guess and test:
Highest Number, 98, 98, Lowest Number
107 - 97 = 10 (meets range requirement)
97 + 98 + 98 + 107 = 400
400/4 = 100 (meets the mean requirement)
Answer:
40
Step-by-step explanation:
were
Answer:
partitions.
Step-by-step explanation:
As we know if we have x number of elements we can arrange them in x! ways.
here we have 100 elements so we can arrange them in 100! ways.
Now if we have exactly three parts and 1,2 and 3 are in different parts then let consider 1 goes to part 1, 2 goes to part 2, and 3 goes to part 3. For the remaining 97 elements, we have three options for each element, thus we can arrange the remaining 97 elements in ways/partitions.
each of these partition will contain only three elements.
