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3 years ago

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The waitress and hostess at a restaurant share the tips at the end of every shift. Between the two of them, they earn an average

of $90.00 in tips. The waitress takes $65.00, and the hostess takes $25.00.
If they continue to earn money at this rate, how much will the waitress receive if they earn $270.00 at the end of the shift?

2 answers:

iris [78.8K]3 years ago

5 0

The waitress will receive $195 in tips te hostess will get 75

boyakko [2]3 years ago

4 0

Answer:

The waitress receive $195 if they earn $270.00 at the end of the shift.

Step-by-step explanation:

The waitress and hostess at a restaurant share the tips at the end of every shift.

They earn an average of $90.00 in tips.

The waitress takes $65.00

Waitress Share fraction = $\frac{65}{90}$

The hostess takes $25.00

Hostess Share fraction = $\frac{25}{90}$

Now we are supposed to find how much will the waitress receive if they earn $270.00 at the end of the shift

So, Waitress Share = $\frac{65}{90} \times 270=195$

Hence The waitress receive $195 if they earn $270.00 at the end of the shift

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Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

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Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

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