Add all the minutes together, then divide by the number of students:
21+35+77+48+24+29+37+98+30+7 = 406 total minutes
10 students.
Average = 406 / 10 = 40.6 minutes per week.
The answer is A.
Answer:
We know angle AOC is a right triangle,
therefor
90 - 73 = x
Angle X is 17.
Answer:
-5.7
Step-by-step explanation:
OUR EQUATION IS A QUADRATIC EQUATION
Let Δ be our dicriminant :
- a= 1
- b= 5
- c= -4
- Δ= 5²-4*1*(-4) =25 +16 =41≥0
- so we have two solutions x and y
- x= (-5-√41)/2 = -5.7
Answer:
<h2>a³-b³ = (a-b)(a²+ab+b²)</h2>
Step-by-step explanation:
let the two perfect cubes be a³ and b³. Factring the difference of these two perfect cubes we have;
a³ - b³
First we need to factorize (a-b)³
(a-b)³ = (a-b) (a-b)²
(a-b)³ = (a-b)(a²-2ab+b²)
(a-b)³ = a³-2a²b+ab²-a²b+2ab²-b³
(a-b)³ = a³-b³-2a²b-a²b+ab²+2ab²
(a-b)³ = a³-b³ - 3a²b+3ab²
(a-b)³ = (a³-b³) -3ab(a-b)
Then we will make a³-b³ the subject of the formula from the resultinh equation;
a³-b³ = (a-b)³+ 3ab(a-b)
a³-b³ = a-b{(a-b)²+3ab}
a³-b³ = a-b{a²+b²-2ab+3ab}
a³-b³ = (a-b)(a²+b²+ab)
a³-b³ = (a-b)(a²+ab+b²)
The long division problem that can be used is (a-b)(a²+ab+b²)
Given:
A box-and-whisker plot of data set.
To find:
The percentage of the data values that are greater than 80.
Solution:
From the given box-and-whisker plot, it is clear that:
Minimum value = 8
First quartile = 60
Median = 68
Third quartile = 80
Maximum value = 92
We know that the 25% the data value are greater than or equal to third quartile because the third quartile divides the data in 75% to 25% and 80 is the third quartile.
Therefore, about 25% of the data values that are greater than 80.
