wrong place
189 - One digit is right and in its
Place
964- Two digits are correct but both
are in the wrong place
523 - All digits are wrong
286 - One digit is right but in the
wrong place
What
are
the three numbers !
The three numbers are 6, 7 and 9
Explanation:
First of all exclude the numbers which aren't correct.
From the 4th data,We can conclude the digit 5, 2 and 3 are wrong.
Now,
We've to find the three digits :
From case 4: 5, 2 and 3 are wrong
So, remove 5, 2 and 3 from all the cases
The numbers are:
From 1st and 3rd we can say that: 4 is present in both the case but wrongly placed.
Now the digits become:
From 3rd and 4th case we can say that: 9 and 6 are correct but wrongly placed, 8 is not the number and 6 is placed at first
So, the two numbers of a 3 digit number are 6 and 9
From 2nd, 3rd and 4th case, the position of 6 and 9 are:
6 _ 9
From 1st case, 7 is the digit but wrongly placed. So, the 3 digit number becomes:
6 7 9
Answer:
5 3/7 or 37/7
Answer:
Quadrants 3 and 4
Step-by-step explanation:
Quadrants 3 &4 represent negative integers. If 0 on the x axis represents the average/mean number of people there, then 3& 4 show the decrease from the average number.
If the listing price for Candy Cane Creek is $864, 550, how much commission would you receive?
Answer:
he received commission amount =7% of $778,095=$54466.65
Step-by-step explanation:
commission =7%
marked price=$778,095
commission amount =7% of $778,095=$54466.65
Answer: $60,518.50
Step-by-step explanation:
864,500 x 0.07
Answer:
V = π (-2 (ln 2)² + 4 ln 2 − 1)
V ≈ 2.55
Step-by-step explanation:
V = π ∫₁² (1 − (ln x)²) dx
V/π = ∫₁² (1 − (ln x)²) dx
V/π = ∫₁² dx − ∫₁² (ln x)² dx
V/π = x |₁² − ∫₁² (ln x)² dx
V/π = 1 − ∫₁² (ln x)² dx
To evaluate the second integral, integrate by parts.
If u = (ln x)², then du = 2 (ln x) / x dx.
If dv = dx, then v = x.
∫ u dv = uv − ∫ v du
= (ln x)² x − ∫ x (2 (ln x) / x) dx
= x (ln x)² − 2 ∫ ln x dx
Integrate by parts again.
If u = ln x, then du = 1/x dx.
If dv = dx, then v = x.
∫ u dv = uv − ∫ v du
= x ln x − ∫ x (1/x dx)
= x ln x − ∫ dx
= x ln x − x
Substitute:
∫ (ln x)² dx = x (ln x)² − 2 ∫ ln x dx
∫ (ln x)² dx = x (ln x)² − 2 (x ln x − x)
∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x
Substitute again:
V/π = 1 − ∫₁² (ln x)² dx
V/π = 1 − (x (ln x)² − 2x ln x + 2x) |₁²
V/π = 1 + (-x (ln x)² + 2x ln x − 2x) |₁²
V/π = 1 + (-2 (ln 2)² + 4 ln 2 − 4) − (-1 (ln 1)² + 2 ln 1 − 2)
V/π = 1 − 2 (ln 2)² + 4 ln 2 − 4 + 2
V/π = -2 (ln 2)² + 4 ln 2 − 1
V = π (-2 (ln 2)² + 4 ln 2 − 1)
V ≈ 2.55