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2 years ago

2 is what percent of 10

Mathematics

1 answer:

kolezko [41]2 years ago

4 0

Answer:

2 is 20% of 10

Step-by-step explanation:

hope this helps

pls rate brainliest

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Help me please :) thank you

Ghella [55]

Answer:

do 10 20 20 40 50

Step-by-step explanation:

hope it helps sorry if it does not if not then try 1 2 3 4 5

6 0

2 years ago

Hese are the values in Paul’s data set.

andrew-mc [135]

Answer: See description below.

The residuals are the differences between the predicted values and the actual values. You will need to make a scatter plot of each difference.

Here are the five points that you will need to plot:
20 - 21 = -1 (1, -1)
17 - 16 = 1 (2, 1)
9 - 10 = -1 (4, -1)
6 - 5 = 1 (5, 1)
2 - 2 = 0 (6, 0)

8 0

3 years ago

Read 2 more answers

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

The ratio of students to teachers at a school is 19:1 how many students are there in each total of people?

Valentin [98]

Answer:

Step-by-step explanation:

19*40= 760

3 0

2 years ago

Read 2 more answers

Pls pls pls pls pls help

raketka [301]

Answer:38x-34

Step-by-step explanation:

f(x)=x^2+3x-7

g(x)=5x-3

We multiply the entire F equation times two, (x^2+3x-7)*4=

(4x^2+12x-28)

Now the entire g equation by 2, (5x-3)*2

(10x-6)

Now we add both equation

(4x^2+12x-28)+(10x-6)

(4x^2+22x-34)

(4x*4x+22x-34)

(16x+22x-34)

38x-34

Hopefully this is correct :)))

7 0

2 years ago