The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is 924, 000 ways
<h3>How to determine the combination</h3>
Note that the formula for combination is given as;
Combination = 
From the information we have that;
There are 28 questions in the pool
The test should have a total of 10 questions;
6 difficult , 10 average and 12 easy questions
We are asked to determine the combination of;
3 difficult questions
4 average questions
3 easy questions
6C3 = 
6C3 = 
6C3 = 20
10C4 = 
10C4 = 
10C4 = 210
12C4 = 
12C4 = 220
The number of ways of constructing the questions is
= 20 × 210 × 220
= 924, 000 ways
Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is 924, 000 ways
Learn more about combination here:
brainly.com/question/4658834
#SPJ1
Answer:
a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle
Step-by-step explanation:
Let l = the original length of the original rectangle
Let w = the original width of the original rectangle
From the description of the problem, we can construct the following two equations
l=2*w (Equation #1)
(l+4)*(w+4)=l*w+88 (Equation #2)
Substitute equation #1 into equation #2
(2w+4)*(w+4)=(2w*w)+88
2w^2+4w+8w+16=2w^2+88
collect like terms on the same side of the equation
2w^2+2w^2 +12w+16-88=0
4w^2+12w-72=0
Since 4 is afactor of each term, divide both sides of the equation by 4
w^2+3w-18=0
The quadratic equation can be factored into (w+6)*(w-3)=0
Therefore w=-6 or w=3
w=-6 can be rejected because the length of a rectangle can't be negative so
w=3 and from equation #1 l=2*w=2*3=6
I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.
What we have so far:
Kinetic energy = 0. The reason behind that is because the beam is not moving at a height of 40m.
Gavity, g = 9.8m/s²
Height = 40m
Potential energy = mgh; this is equal to 0 because m, stands for mass and in this problem, we do not have a value for the mass of the beam. Hence, 0 x 9.8m/s² x 40m = 0. Potential energy = 0.
Solution:
We will use the equation of Total energy:
TE = potential energy + kinetic energy
TE = 0 + 0
∴ TE = 0
The answer is: Assuming no air resistance, the total energy of the beam as it hits the ground is 0.
Answer:
-0.9
Step-by-step explanation:
First we have to distribute. -1 times t is -1t and -1 times 8 is -8. Now, our equation will look like this: -1t - 8 = -7.1
Next, we have to carry our -8 to the other side of the euqation. So, we have to plus 8 on both sides. -7.1 plus 8 is 0.9. Now, our equation will look like this,
-1t = 0.9.
Lastly, we have to divide -1 from both sides. 0.9 divided by -1 is -0.9. So therefore, t = -0.9.