Answer:
30,240 ways
Step-by-step explanation:
This question is bothered on permutation. Permutation has to do with arrangement.
If there are 10 computers and 5 students, the number of ways students will sit at the computers if no computer has more than one student can be expressed as;
10P5 = 10!/(10-5)!
10P5 = 10!/(5)!
10P5 = 10*9*8*7*6*5!/5!
10P5 = 10*9*8*7*6
10P5 = 30,240
Hence the number of ways is 30,240 ways
Answer:
-19
Step-by-step explanation:
2×2-2z4+y2-x2+z4
now we put the value,
4-2(2to the power 4)+(3 to the power 2)-(-4 to the power 2)+(2 to the power 4)
4-2×16+9-16+16
4-32+9
4+9-32
13-32
-19
please check the answer again i am sorry if it's wrong
Answer:
Option D.The ratio of the radius of a circle to its circumference
Step-by-step explanation:
<em>Verify each statement</em>
A) The ratio of the circumference of a circle to its diameter
The statement is True
Because
so
B) Approximated to be 3.14
The statement is True
Because 
C) Approximated to be 22/7
The statement is True
Because
D) The ratio of the radius of a circle to its circumference
The statement is not True
Because
Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.
We are to find the probability of selecting 1 red apple and 2 yellow apples.
Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.
Then, we have
Therefore, the probability of event A is given by
Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
By algebra properties we find the following relationships between each pair of algebraic expressions:
- First equation: Case 4
- Second equation: Case 1
- Third equation: Case 2
- Fourth equation: Case 5
- Fifth equation: Case 3
<h3>How to determine pairs of equivalent equations</h3>
In this we must determine the equivalent algebraic expression related to given expressions, this can be done by applying algebra properties on equations from the second column until equivalent expression is found. Now we proceed to find for each case:
First equation
(7 - 2 · x) + (3 · x - 11)
(7 - 11) + (- 2 · x + 3 · x)
- 4 + (- 2 + 3) · x
- 4 + (1) · x
- 4 + (5 - 4) · x
- 4 - 4 · x + 5 · x
- 4 · (x + 1) + 5 · x → Case 4
Second equation
- 7 + 6 · x - 4 · x + 3
(6 · x - 4 · x) + (- 7 + 3)
(6 - 4) · x - 4
2 · x - 4
2 · (x - 2) → Case 1
Third equation
9 · x - 2 · (3 · x - 3)
9 · x - 6 · x + 6
3 · x + 6
(2 + 1) · x + (14 - 8)
[1 - (- 2)] · x + (14 - 8)
(x + 14) - (8 - 2 · x) → Case 2
Fourth equation
- 3 · x + 6 + 4 · x
x + 6
(5 - 4) · x + (7 - 1)
(7 + 5 · x) + (- 4 · x - 1) → Case 5
Fifth equation
- 2 · x + 9 + 5 · x + 6
3 · x + 15
3 · (x + 5) → Case 3
To learn more on algebraic equations: brainly.com/question/24875240
#SPJ1
