All categories

3 years ago

What is the average rate of change for this exponential function for the

1 answer:

7 0

What is the average rate of change for this exponential function for the interval from x=0 to x=2
m = (f(2) - f(0)) / (2 - 0)
m = (4 - 1) / (2 - 0)
m = 3 / 2
m = 1.5

You might be interested in

Х = бу - 1<br> бу + 5х = 31

Kobotan [32]

Answer:

look search math-way and type your problem in .

Step-by-step explanation:

5 0

3 years ago

How to solve the polynomial 5x+3y+6x+9y

ANEK [815]

5x+3y+6x+9y

All you'd need to do is combine like terms. So add 5x to 6x and add 3y to 9y.

(5x+6x)+(3y+9y)

Final Answer: 11x + 12y

3 0

3 years ago

Read 2 more answers

At prom, Dina asks a boy to dance. A moment later, her friend decides to ask a boy to dance, too.

leonid [27]

Answer:

Dina's friend is jealous of Dina's confidence so she wanted to "what up" her :)

Step-by-step explanation:

3 0

3 years ago

PLSSSS HELP MEEEE I NEED TO TURN THIS IN BY THE END OF CLASS PLSSS SOMEONE HELP ME WITH THIS!!!!!!

Annette [7]

Answer:

1) 10

2) 22

3) 28

4) 17.5

5) 22

6) 13.5

7) 19.5

8) 11

9) 7.5

4 0

3 years ago

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

3 years ago