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Natali5045456 [20]
3 years ago
12

Type the correct answer in the box.

Mathematics
1 answer:
zhenek [66]3 years ago
8 0

Using Excel formulas

 =LARGE(A1:A20, 1)/COUNT (A1:A20)

 Count (Ai:An) Function that allows to count the total number of values in a cells range between Ai and An that contain only numbers.

 Large (Ai:An, k) Function that allows to find the value of the largest number in a cells range between Ai and An, the number k determine the k-th largest value

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Can someone help me? you might get a crown
Step2247 [10]

Answer:

2.1 mph

Step-by-step explanation:

7 0
3 years ago
Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.
elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}

Consider a vector field F=x\^i+y\^j+z\^k

Then the divergence of the vector F is,

div F = \nabla.F = (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)

and the curl of the vector F is,

curl F = \nabla\times F = \^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})

<h3>Calculation:</h3>

The given vector fields are:

F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k

1) Verifying the identity: \nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

Consider L.H.S

⇒ \nabla.(aF1+bF2)

⇒ \nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))

⇒ \nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the dot product between these two vectors,

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(1)

Consider R.H.S

⇒ a\nabla.F1+b\nabla.F2

So,

\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)

⇒ \nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}

\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)

⇒ \nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}

Then,

a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(2)

From (1) and (2),

\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

2) Verifying the identity: \nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Consider L.H.S

⇒ \nabla\times(aF1+bF2)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the cross product,

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y}) ...(3)

Consider R.H.S,

⇒ a\nabla\times F1+b\nabla\times F2

So,

a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))

⇒ \^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})

a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))

⇒ \^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})

Then,

a\nabla\times F1+b\nabla\times F2 =

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})

...(4)

Thus, from (3) and (4),

\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

8 0
1 year ago
Diego's club earns money for charity when members of the club perform community service after school,each student who does commu
Arlecino [84]

Answer:

No

Step-by-step explanation:

Given that:

Amount earned when a student does comminty service = $5

Number of students in club = 12

Amount earned E, is a function of number of members (n) who does community service.

Hence,

E = n * amount earned per member

Is 24 a possible output?

To test, then E = 24

24 = n * 5

24 = 5n

24/5 = 5n/5

4.8 = 5

Hence, 24 is not a possible output, be use number of membwra cannot be a decimal

4 0
3 years ago
Of the relation
RideAnS [48]

Answer:

Domain is x and Range is y

Step-by-step explanation:

Domain-(16,-15,22,-1,-6)

Range-(5,-17,-16,-4,-14)

5 0
3 years ago
Almost all companies utilize some type of year-end performance review for their employees. Human Resource (HR) at a university's
balu736 [363]

Answer:

The Answer is 76.

Step-by-step explanation:

Given the normal distribution " 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable'',  we can see that exemplary employees are top 10% rated employees.

We have the formula for normal distribution:

z=(X-M)÷σ

where z is the <em>minimum z-score </em>for top 10% employee, X is the <em>minimum </em>score for top 10% employee, M is the <em>mean</em> of the score distribution, σ is the <em>standard deviation</em> of the score distribution.

The z-score we are looking for is the value "a" that separates the highest 10% from the lowest 90% i.e. P(z≤a)=0.90

If we look at z-table, corresponding value for a is 1.28155

We can now put the values in the formula:

1.28155=\frac{X-50}{20}

So X=(1.28155×20)+50=75.631

Therefore minimum score for exemplary employee is 76.

3 0
2 years ago
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