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3 years ago

How do you write 25/20 as a percentage

1 answer:

8 0

$\dfrac{25}{20}=\dfrac{25\cdot5}{20\cdot5}=\dfrac{125}{100}=1\dfrac{25}{100}=1.25=1.25\cdot100\%=\boxed{125\%}\\\\\dfrac{25}{20}=\dfrac{25}{20}\cdot100\%=25\cdot5\%=\boxed{125\%}$

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A firm producing 7 units of output has an average total cost of Rs. 150 and has to pay Rs.350 to its fixed factors of production

maria [59]

Rs 100 of the average total cost is made up of variable costs.

Step-by-step explanation:

Given:

Number of output the firm produces= 7 units

Average cost of the output= Rs. 150

fixed factors of production = Rs.350

To Find:

How much of the average total cost is made up of variable costs=?

Solution:

we know that,

Average total cost= total cost/ number of output units produced

substituting the values, we get

$150=\frac{\text{Total cost}}{7}$

Total cost= 1050

we know that Total fixed cost = 350

Total cost = Total fixed cost + Total variable cost

plug in the known values.

1050= 350 + Total variable cost

Total variable cost = 1050-350

Total variable cost =700

For 7th unit $\frac{700}{7}$ = 100

5 0

3 years ago

Rewrite 4 6/7 as an improper fraction.<br><br> sorry this is the correct question

goblinko [34]

The correct answer is, obviously, 34/7.

8 0

3 years ago

Read 2 more answers

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$