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2 years ago

P-4=-9+p i don't know how to determine whether its an infinite solution, or n solution.

Mathematics

2 answers:

lara [203]2 years ago

6 0

Subtract P from both sides and be left with -4 = -9, which means no solutions.

adoni [48]2 years ago

4 0

♥ Solve:
Subtract "P" from both sides.
Answer:
<span>-4 = -9
</span>Which means
No solutions.
Final answer: No solutions.

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Ardem collected data from a class survey. He then randomly selected samples of five responses to generate four samples.

scoundrel [369]

Solution: The sample mean of sample 1 is:

$\bar{x}=\frac{4+5+2+4+3}{5}= \frac{18}{5}=3.6$

The sample mean of sample 2 is:

$\bar{x}=\frac{2+2+6+5+7}{5}= \frac{22}{5}=4.4$

The sample mean of sample 3 is:

$\bar{x}=\frac{4+6+3+4+1}{5}= \frac{18}{5}=3.6$

The sample mean of sample 4 is:

$\bar{x}=\frac{5+2+4+3+6}{5}= \frac{20}{5}=4$

The minimum sample mean of the four sample means is 3.6 and maximum sample mean of the four sample means is 4.4.

Therefore, using his four samples, between 3.6 and 4.4 will Ardem's actual population mean lie.

Hence the option 3.6 and 4.4 is correct

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3 years ago

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A Create a linear expression

frez [133]

Answer:

(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB. Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .Step-by-step explanation:

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2 years ago

Find parametric equations for the sphere centered at the origin and with radius 3. Use the parameters s and t in your answer.

sdas [7]

Radius, r = 3

The equation of a sphere entered at the origin in cartesian coordinates is

x^2 + y^2 + z^2 = r^2

That in spherical coordinates is:

x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)

where you can make u = r cos(phi) to obtain the parametrical equations

x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u

where theta goes from 0 to 2π and u goes from -r to r.

In our case r = 3, so the parametrical equations are:

Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u

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3 years ago

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n200080 [17]

The awnser is (-1+3/x-3/x2+1/x3 )

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2 years ago

A square of side length s lies in a plane perpendicular to a line L. One vertex of the square lies on L. As this square moves a

user100 [1]

Answer:

Part (A) The required volume of the column is $s^2h$ .

Part (B) The volume be $s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}$ .

Step-by-step explanation:

Consider the provided information.

It is given that the we have a square with side length "s" lies in a plane perpendicular to a line L.

Also One vertex of the square lies on L.

Part (A)

Suppose there is a square piece of a paper which is attached with a wire through one corner. As you blow it up it spins around on the wire.

This square moves a distance h along L, and generate a corkscrew-like column with square.

The cross section will remain the same.

So the cross section area of original column and the cross section area of twisted column at each point will be the same.

The volume of the column is the area of square times the height.

This can be written as:

$s^2h$

Hence, the required volume of the column is $s^2h$ .

Part (B) What will the volume be if the square turns twice instead of once?

If the square turns twice instead of once then the volume will remains the same but divide the volume into two equal part.

$s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}$

Hence, the volume be $s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}$ .

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3 years ago