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bagirrra123 [75]
2 years ago
6

The sequence an = one third (3)n - 1 is graphed below: coordinate plane showing the points 2, 1; 3, 3; and 4, 9 Find the average

rate of change between n = 3 and n = 4. (6 points) negative one sixth one sixth −6 6
Mathematics
2 answers:
charle [14.2K]2 years ago
7 0

9514 1404 393

Answer:

  6

Step-by-step explanation:

The slope of the line between (3, 3) and (4, 9) is given by the slope formula:

  m = <average rate of change> = (y2 -y1)/(x2 -x1)

  m = (9 -3)/(4 -3) = 6/1

  m = 6

The average rate of change on the interval [3, 4] is 6.

Alborosie2 years ago
4 0

Answer:

the answer is 6

Step-by-step explanation:

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The probability of being 25-35 years and having a haemoglobin level above 11 is 34%.

The probability of having a haemoglobin level above 11 is 36%.

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Probability determines the odds that a random event would occur. The odds of the event happening lie between 0 and 1.

The probability of being 25-35 years and having a haemoglobin level above 11 = number of people between 25 - 35 that have a level above 11 / total number of people between 25 - 35

44 / 128 = 34%

The probability of having a haemoglobin level above 11  = number of people with a level above 11 / total number of respondents

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Read 2 more answers
Questions Below. Would Appreciate Help!
kherson [118]

Answer:

The function that could be the function described is;

f(x) = -10 \cdot cos \left (\dfrac{2 \cdot \pi }{3} \cdot x \right ) + 10

Step-by-step explanation:

The given parameters of the cosine function are;

The period of the cosine function = 3

The maximum value of the cosine function = 20

The minimum value of the cosine function = 0

The general form of the cosine function is presented as follows;

y = A·cos(ω·x - ∅) + k

Where;

\left | A \right | = The amplitude = Constant

The period, T = 2·π/ω

The phase shift, = ∅/ω

k = The vertical translation = Constant

Therefore, by comparison, we have;

T = 3 = 2·π/ω

∴ ω = 2·π/3

The range of value of the cosine of an angle are;

-1 ≤ cos(θ) ≤ 1

Therefore, when A = 10, cos(ω·x - ∅) = 1 (maximum value of cos(θ)) and k = 10, we have;

y = A × cos(ω·x - ∅) + k

y = 10 × 1 + 10 = 20 = The maximum value of the function

Similarly, when A = 10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, we get;

y = 10 × -1 + 10 = 0 = The minimum value of the function

Given that the function is a reflection of the parent function, we can have;

A = -10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, to get;

y = -10 × -1 + 10 = 20 = The maximum value of the function

Similarly, for cos(ω·x - ∅) = 1 we get;

y = -10 × 1 + 10 = 0 = The minimum value of the function

Therefore, the likely values of the function are therefore;

A = -10, k = 10

The function is therefore presented as follows;

y = -10 × cos(2·π/3·x) + 10

8 0
2 years ago
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