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enot [183]
3 years ago
7

If your algorithm runs its critical section 1 + 2 + 3 + ... + (n-2) + (n-1) + n times, what is the asymptotic behavior of the al

gorithm?
Mathematics
1 answer:
Paraphin [41]3 years ago
4 0
If 1 and 2 becomes multiplayed gave 3 and 3+3=4 (4-2)+(4-1)= 2+3=4 and 4 is n.
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C is the midpoint of AD. B is the midpoint of AC. BC = 12. What is the length of AD
kvasek [131]
48 I would believe is the answer
6 0
3 years ago
Read 2 more answers
F(x)=2x-3 g(x)=x^2+1 find g(2)
Hitman42 [59]

Answer:

g(2)=5

Step-by-step explanation:

since g(x) is equal to x^2+1, you just plug in the value of 2 for x, to get 2^2+1, and 2 to the power of 2 (also known as 2 squared) is equal to 4, which makes your equation 4+1, which equals 5.

3 0
3 years ago
Can u help me with a math question plz?
Furkat [3]

Given: 48 students took the exam.

A total of 18 + 10 = 28 students scored 80% or above.

To find the percentage, take the part (28 students) divided by the whole (48 students) then multiply the quotient by 100.

(28 students / 48 students) x 100 = 58.33%

58.33% of the students who took the exam scored an 80% or higher

6 0
3 years ago
A placebo is a substance that should elicit no positive response in an experimental subject.
adelina 88 [10]

Answer:

  A)  True

Step-by-step explanation:

In an experiment that has the purpose of testing the efficacy of a procedure or drug, comparison is made against the efficacy of a placebo, a procedure or drug that is <em>intended to have no effect whatever</em>.

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Famously, a placebo is often found to be nearly as effective (or even more effective) than the procedure or drug on trial. This effect is known as "the placebo effect."

7 0
3 years ago
Find the area between the two functions
Inga [223]

The area between the two functions is 0

<h3>How to determine the area?</h3>

The functions are given as:

f₁(x)= 1

f₂(x) = |x - 2|

x ∈ [0, 4]

The area between the functions is

A = ∫[f₂(x) - f₁(x) ] dx

The above integral becomes

A = ∫|x - 2| - 1 dx (0 to 4)

When the above is integrated, we have:

A = [(|x - 2|(x - 2))/2 - x] (0 to 4)

Expand the above integral

A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]

This gives

A = [2 - 4] - [-2- 0]

Evaluate the expression

A = 0

Hence, the area between the two functions is 0

Read more about areas at:

brainly.com/question/14115342

#SPJ1

5 0
1 year ago
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