The Roosevelts and the Jaspers live in the same city and pay the same sales

Mathematics

2 answers:

Nana76 [90]3 years ago

7 0

Answer:

Yes, because the Jaspers paid a higher percentage of their income in sales tax than the Roosevelts did.

Step-by-step explanation:

Regressive tax is a constant, instead of a percentage of income. Those with a lower income would pay a higher percentage of their income. Both families, irregardless of income, paid $16,000 in taxes.

APEX

Amanda [17]3 years ago

7 0

Answer:

Yes, because the Jaspers paid a higher percentage of their income in sales tax than the Roosevelts did.

Step-by-step explanation:

A regressive tax is one that collects a smaller percentage of the income as the person earns more. That is to say that at higher profit or higher income, the percentage of taxes that must be paid on the total tax base is lower.

In this way the poor are relatively more affected than the rich.

The Roosevelts made $91.000 (high income)

The Jaspers made $37.000 (low income)

But both paid the same sales tax rate of $16,000.

This type of tax does not have a redistribution effect of wealth. On the contrary, if it is a very high tax, it can accentuate inequality in a society.

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F ind the volume under the paraboloid z=9(x2+y2) above the triangle on xy-plane enclosed by the lines x=0, y=2, y=x

olga nikolaevna [1]

Answer:

The answer is 48 units³

Step-by-step explanation:

If we simply draw out the region on the x-y plane enclosed between these lines we realize that,if we evaluate the integral the limits all in all cannot be constants since one side of the triangular region is slanted whose equation is given by y=x. So the one of the limit of one of the integrals in the double integral we need to evaluate must be a variable. We choose x part of the integral to have a variable limit, we could well have chosen y's limits as non constant, but it wouldn't make any difference. So the double integral we need to evaluate is given by,

$V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {z} \, dx dy\\V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {9(x^{2}+y^{2})} \, dx dy$

Please note that the order of integration is very important here.We cannot evaluate an integral with variable limit last, we have to evaluate it first.after performing the elementary x integral we get,

$V=9\int\limits^2_0 {4y^{3}/3} \, dy$