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lesya [120]
3 years ago
11

(where it says "lve" at the top it's says solve)​

Mathematics
1 answer:
raketka [301]3 years ago
7 0

Answer:

a

Step-by-step explanation:

its a fact 3x3x3=27

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PLEASE ANSWER ASAP<br> ..............................................
Tema [17]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
Calculus AB Assignment: Writing Symmetrical Functions 4. The domain of f(x) is [O, ..) and 1 2 3 4 f(x) { } 1o in Х 1 2 1 5 1 17
AveGali [126]

Answer:

Step-by-step explanation:

5 0
2 years ago
Which equation should be used to calculate the 43rd partial sum for the arithmetic sequence.
MariettaO [177]

Answer: First Option : Sₙ= n/2(a₁ + aₙ)

Step-by-step explanation:

The nth partial sum of an arithmetic sequence or the sum of the first n terms of the arithmetic series can be defined as the sum of a finite number of term in an arithmetic sequence.

It is calculated using the formula:

Sₙ= n/2(a₁ + aₙ)

Where :

a₁ = First term

aₙ = last term

n = number of terms

6 0
2 years ago
which shows the students in order from greatest test score to lease test score Alex 0.95 Octavia 16/20 Tonya 9/10 Wilson 0.87 ​
Nonamiya [84]

Answer:

Alex (0.95), Tonya (0.90), Wilson (0.87), Octavia (0.80)

Step-by-step explanation:

To solve this, we have to convert all of the scores into either a decimal or a fraction. In this case, I will be solving all of them into decimals since it is easier.

Alex: 0.95

Octavia: 16/20 = 0.80

Tonya: 9/10 = 0.90

Wilson: 0.87

Therefore, we can order these from greatest score to least score.

Alex (0.95), Tonya (0.90), Wilson (0.87), Octavia (0.80)

4 0
2 years ago
Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t​
juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.

 

6 0
3 years ago
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