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3 years ago

A DVD player costs $250 and TV costs $600. With a 19% commission rate, how much

Mathematics

1 answer:

Vika [28.1K]3 years ago

4 0

In this question, we're trying to find how much a sale clerk earns from a sale.

We know that the DVD player costs $250

We also know that the TV costs $600

The sales clerk has a 19% commission rate, this is the money that they make from the sale.

To find the money the sales clerk got from the sale, we have to add the total spending and multiply it by 0.19

250 + 600 = 850

850 · 0.19 = 161.5‬

This means that the sales clerk made $161.5‬0 in commission from the sale.

Answer:

$161.50‬

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How would I solve 1 and 2?

Rus_ich [418]

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2 years ago

Sonja's credit card has an APR of 21%, calculated on the previous monthly

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2 years ago

In one school, a half of all students who like math like science as well. Also, in that school, a third of all students who like

Marizza181 [45]

Answer: 2/3

Step-by-step explanation:

N is the total number of students

M is the number of students thta like math

S is the number of students that like science.

We know that half of the elements in M also are elements from S

And a third of the elements of S also are elements of M

And because those elements are common elements for both sets, we should have that:

M/2 = S/3

then we have that:

M = (2/3)*S

The ratio is 2/3

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3 years ago

Read 2 more answers

HELP ME ASAP YOU GET 10 POINTS

snow_lady [41]

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2 years ago

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

2 years ago