The product of 2x^3 and 6x^5 is

If N = f(x) and f(x) = 3x^2 −x−5

harina [27]

Answer:

I really don't know so sorry

8 0

3 years ago

Y=4x+28 sorry some is cut off

Rus_ich [418]

Add
4
x
4
x
to both sides of the equation.
y
=
−
28
+
4
x
y
=
-
28
+
4
x
Rewrite in slope-intercept form.
Tap for more steps...
y
=
4
x
−
28
y
=
4
x
-
28
Use the slope-intercept form to find the slope and y-intercept.
Tap for more steps...
Slope:
4
4
y-intercept:
−
28
-
28
Any line can be graphed using two points. Select two
x
x
values, and plug them into the equation to find the corresponding
y
y
values.
Tap for more steps...
x
y
2
−
20
3
−
16
x y 2 -20 3 -16
Graph the line using the slope and the y-intercept, or the points.
Slope:
4
4
y-intercept:
−
28
-
28
x
y
2
−
20
3
−
16
x y 2 -20 3 -16
image of graph
y
−
4
x
=
−
2
8
y
-
4
x
=
-
2
8

28
x
28
x
28
x
2
28
x
2
28
x
3
28
x
3

7 0

3 years ago

Which of the following is solution to the compound inequality below?

djverab [1.8K]

Solve compound inequalities in the form of or and express the solution graphically. ... determining thesolution to the compound inequality, as in the example below. .... Which of the following compound inequalities represents the graph

4 0

2 years ago

Read 2 more answers

A person who weighs 120 pounds on Earth weighs 20 pounds on the Moon. How much does a 93-pound person weigh on the Moon?

Debora [2.8K]

93 * 20 / 120 = 15.5

a 93-pound person weighs on the Earth will weigh 15.5 lbs on Moon

8 0

3 years ago

Read 2 more answers

Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir

lys-0071 [83]

Answer:

a) $\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

b) $Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

c) $\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

If we apply this formula for the four methods we got:

$\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

$Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

$\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

3 0

2 years ago