We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:
We have then a total number of grades of 39.
The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.
Then, for grade A, the relative frequency (RF) will be:
This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.
Now, to complete the table we do the same for the other frequencies:
For grade B:
For grade C:
For grade D:
For grade F:
Answer:
Step-by-step explanation:
((95.26 - 78.17) / 78.17)x100
21.86260714% percentage decrease
The phone call was worth his time
To solve this problem, we need to get the variable x alone on one side of the equation. To begin, we are going to use the distributive property twice on the left side of the equation to expand the multiplication and get rid of the parentheses.
4(x-1) - 2(3x + 5) = -3x -1
4x - 4 -6x - 10 = -3x - 1
Next, we should combine like terms on the left side of the equation. This means we should add/subtract the variable terms and the constant terms in order to simplify this equation further.
-2x - 14 = -3x - 1
Then, we have to add 3x to both sides of the equation to get the variable terms all on the left side of the equation.
x - 14 = -1
After that, we should add 14 to both sides of the equation to get the variable x alone one the left side of the equation.
x = 13
Therefore, the answer is 13.
Hope this helps!
Answer:
Your answer is: Infinitely Many
Any value of x makes the equation true. Also, they are all real numbers.
Step-by-step explanation:
If you are solving to find all Complex Number Solutions: ↓
Solve the equation to find all of the complex solutions and that means that equation will be always true.
Hope this helped : )
Try this solution:
1. Note, that 100 is divisible by 4, and 999 is not divisible by it, only 996. This is an arithmetic sequence.
2. a1;a2;a3;a4;...a(n) the sequence, where a1=100; a2=104; a3=108; a4=112; ... etc., and a(n)=996. n=?
3. using a formula for n-term of the sequence: a(n)=a1+d(n-1), where a(n)=996; a1=100 and d=4 (according to the condition ' is divisible by 4'). Then 100+4(n-1)=996; ⇒ 4n=900; ⇒ n=225 (including 100).
answer: 225
