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3 years ago

13

Which function could be a stretch of the exponential decay function shown on the graph?

2 answers:

BartSMP [9]3 years ago

6 0

Answer:

itd b on edg

Step-by-step explanation:

dezoksy [38]3 years ago

4 0

Answer:

f(x) = $(\frac {1}{2})^{(6x)}$

Step-by-step explanation:

The given options are,

a) f(x) = $2^{(6x)}$

b) f(x) = $(\frac {1}{2})^{(6x)}$

c) f(x) = $2 \times (\frac {1}{6})^{x}$

d) f(x) = $\frac {1}{2} \times (\frac {1}{6})^{x}$

Now, clearly a) is a monotonically increasing function, hence discarded, and both of c) and d) don't pass through (0, 1) hence they are also discarded.

Only b) is a decay function which does also pass through (0, 1), hence, b) is the correct option.

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What is the answer for this Problem?

geniusboy [140]

Answer:

Increasing on the interval(s) $-2\leq x\leq -4$

Decreasing on the interval(s) $-1\leq y\leq 1$

Step-by-step explanation:

Anytime the graph is going upward, the function is increasing. However, anytime the graph goes down, the function is decreasing. Look at the image below for further reference. Also, when the function is increasing, the slope is positive, and when the function is decreasing the slope is negative.

The increasing interval of the graph is -3

The decreasing interval of the graph is 0.5

Since the values are in the middle of the interval, it automatically becomes the answer.

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Step-by-step explanation:

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swat32

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago