Find the slope. A 2 B -1/2 C -2 D 1/2

zubka84 [21]

In a boxplot, there are 5 figures that are denoted, these are known as the 5-figure summary:
(1) the black dot to the very left of the boxplot is the minimum, or in other words the lowest data value
(2) the second black dot from the left is Q1 (quartile 1) and the area between the minimum and Q1 represents 25% of the data
(3) the middle dot is the median (or sometimes called Q2), which is the middle value of the data set - in this case it is 24
(4) the black dot to the right of the median represents Q3
(5) the black dot to the very right is the maximum value, or the highest data value

7 0

3 years ago

a proportion relationship is represented by the equation 2x equals 18 y. If y equals kx where K is the constant of proportionali

Mkey [24]

18y=2x
y=1/9 *x
y=k*x
so k=1/9

5 0

4 years ago

Does the following infinite series converge or diverge?

Vladimir79 [104]

The nth term is (presuming the denominator on the first is a typo)

$-\dfrac{n}{(-3)^n}$

The ratio of successive terms is

$\dfrac{ \frac{n+1}{(-3)^(n+1)}}{\frac{n}{(-3)^n}} = \frac{n+1}{-3n}$

The limit is -1/3, absolute value less than 1, so the series converges. It has a sum -- that's what converges means.

7 0

4 years ago

PLEASE HELP: Which is a Linear function? Y=x^3 + 1 5/6x=y-4 Y=8x^2. Y=1/5x

azamat

Y=1/5x is a linear function

4 0

4 years ago

A biologist is studying the growth of a particular species of algae. She writes the fall inclusion to show the radius of the alg

horrorfan [7]

Answer:

Part A)

A reasonable domain to plot the growth function is:

$0\leq d \leq 12$

Part B)

The y-intercept represents that when the biologist started her study, the radius of the algae was five millimeters.

Part C)

The average rate of change from d = 4 to d = 11 was about 0.11. This means that from the 4th day to the 11th, the radius of the algae grew, on average, at a rate of 0.11 mm per day.

Step-by-step explanation:

The radius of the algae f(d) in millimeters after d days is given by the function:

$f(d)=5(1.02)^d$

Part A)

We know that the radius of the algae was approximately 6.34 mm when the biologist concluded her study. To find the reasonable domain, we can substitute 6.34 for f(d) and solve for d. Therefore:

$6.34=5(1.02)^d$

Divide both sides by five:

$(1.02)^d=1.268$

Take the log of both sides with base 1.02:

$\displaystyle d=\log_{1.02}1.268$

Using the Change of Base Property, evaluate for d:

$\displaystyle d=\frac{\log 1.268}{\log 1.02}=11.9903...\approx 12$

So, the biologist concluded her study after 12 days.

Therefore, a reasonable domain to plot the growth function is:

$0\leq d\leq 12$

Part 2)

The y-intercept of the function is when d = 0. Find the y-intercept:

$f(0)=5(1.02)^{(0)}=5(1)=5$

Since d represent the amount of days after the study had begun, the y-intercept represents the radius of the algae on the initial day.

So, when the biologist started her study, the radius of the algae was five millimeters.

Part 3)

To find the average rate of change for a nonlinear function, we find the slope between the two endpoints on the interval.

We want to find the average rate of change of f(d) from d = 4 to d = 11.

Find the endpoints:

$f(4)=5.4121...\text{ and } f(11)=6.2168...$

And find the slope between them:

$\displaystyle m=\frac{f(11)-f(4)}{11-4}=0.1149...\approx 0.11$

Since f(d) measures millimeters and d measures days, this tells us that, on average, the radius of the algae grew by about 0.11 mm per day from the 4th day to the 11th day.

5 0

3 years ago