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Vitek1552 [10]
3 years ago
8

We _____ when calculating the average deviation, because the sum of all deviation scores around the mean equals zero.

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
6 0
Subtract because it has to deal with the many signs
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Dont skip plz i need yall help
neonofarm [45]

Answer: B) 1 and -3

Step-by-step explanation:

On a Quadrant Plane:

Q1: (+,+)

Q2: (-,+)

Q3: (-,-)

Q4: (+,-)

5 0
3 years ago
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In at least 4 complete sentences explain what a reciprocal is in relation to fraction? and how you would find the reciprocal of
sveta [45]

To find the reciprocal of a fraction, you switch the numerator and denominator, for example the reciprocal of 2/5 is 5/2. With our mixed number, we convert to an improper fraction, 17/3Then we find the reciprocal, which is 3/17. I hope this helps!

3 0
3 years ago
WILL GIVE BRAINLIEST IF CORRECT!
kodGreya [7K]

Answer:

1/24

Step-by-step explanation:

1/6 / 4

1/6 x 1/4

1/24

7 0
3 years ago
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What are exponents??????​
xxTIMURxx [149]

Answer:

a quantity representing the power to which a given number or expression is to be raised, usually expressed as a raised symbol beside the number or expression (e.g. 3 in 23 = 2 × 2 × 2).

6 0
2 years ago
Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with
skad [1K]

Parameterize S by the vector function

\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle

The integral of the field over S is then

\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv

\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv

\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}

8 0
2 years ago
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