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4 years ago

Professor Perez used a spreadsheet to compile the data shown. Which value best approximates the

Mathematics

2 answers:

Inessa [10]4 years ago

6 0

Answer:

0.9

Step-by-step explanation:

marishachu [46]4 years ago

3 0

Answer:

The answer is "The value to this question is positive".

Step-by-step explanation:

In this question, the value of the correlation coefficient is lies between -1 to 1.

The value of n is given that is 7, and in this question one value is extra given, that is "64".

To calculate $x^2 = x*x , \ \ \ y^2= y*y\\$ .

please find the attachment.

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(Unitary Method)

Sati [7]

Given :-

12 workers can do a piece of work in 20 days .

To Find :-

How many workers should be added to complete the work in 16 days ?

Solution :-

According to the question ,

→ In 20days 12 workers can do a piece of work.

→ In 1day 12*20 workers can do that work

( Less days , more workers )

→ In 16 days 12*20/16 = 15 workers can do the work.

So 15 -12 = 3 workers should be added to complete the work.

Hence the required answer is 3 .

I hope this helps.

7 0

2 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the

timofeeve [1]

Answer:

152 units²

Step-by-step explanation:

4×8 + 2[(6×8) + (½×6×4)]

= 152

6 0

3 years ago

Read 2 more answers

I need help asap :):):):)

Georgia [21]

Answer:

Step-by-step explanation:

3 0

4 years ago

If a club has 20 members and

DIA [1.3K]

Wiseman’s Durango Guam Cincinnati Franco so

5 0

3 years ago

Read 2 more answers

Use a triple integral to find the volume of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0

Tanzania [10]

Answer:

Volume of the Tetrahedron T = $\frac{1}{3}$

Step-by-step explanation:

As given, The tetrahedron T is bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0

We have,

z = 0 and x + 2y + z = 2

⇒ z = 2 - x - 2y

∴ The limits of z are :

0 ≤ z ≤ 2 - x - 2y

Now, in the xy- plane , the equations becomes

x + 2y = 2 , x = 2y , x = 0 ( As in xy- plane , z = 0)

Firstly , we find the intersection between the lines x = 2y and x + 2y = 2

∴ we get

2y + 2y = 2

⇒4y = 2

⇒y = $\frac{2}{4} = \frac{1}{2}$ = 0.5

⇒x = 2( $\frac{1}{2}$ ) = 1

So, the intersection point is ( 1, 0.5)

As we have x = 0 and x = 1

∴ The limits of x are :

0 ≤ x ≤ 1

Also,

x = 2y

⇒y = $\frac{x}{2}$

and x + 2y = 2

⇒2y = 2 - x

⇒y = 1 - $\frac{x}{2}$

∴ The limits of y are :

$\frac{x}{2}$ ≤ y ≤ 1 - $\frac{x}{2}$

So, we get

Volume = $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}\int\limits^{2-x-2y}_{z=0} {dz} \, dy \, dx$

= $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{[z]}\limits^{2-x-2y}_0 {} \, \, dy \, dx$

= $\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{(2-x-2y)} \, \, dy \, dx$

= $\int\limits^1_0 {[2y-xy-y^{2} ]}\limits^{1-\frac{x}{2}} _{\frac{x}{2} } {} \, \, dx$

= $\int\limits^1_0 {[2(1-\frac{x}{2} - \frac{x}{2}) -x(1-\frac{x}{2} - \frac{x}{2}) -(1-\frac{x}{2}) ^{2} + (\frac{x}{2} )^{2} ] {} \, \, dx$

= $\int\limits^1_0 {(1 - 2x + x^{2} )} \, \, dx$

= ${(x - x^{2} + \frac{x^{3}}{3} )}\limits^1_0$

= 1 - 1² + $\frac{1^{3} }{3}$ - 0 + 0 - 0

= 1 - 1 + $\frac{1 }{3}$ = $\frac{1}{3}$

So, we get

Volume = $\frac{1}{3}$

7 0

3 years ago