Can you please help me with this i will give you a Brainiest

Guido is a citizen and resident of Belgium. He has a full-time job in Belgium and has lived there with his family for the past 1

Korvikt [17]

Answer:

Guido stayed in US in 2018 for 180 days which are greater than 31 days.

Guido Stayed in US in 2018 and in 2017 = (180 + 66) > 183 days.

So, yes Guido does meet US Statutory definition in 2018 and stayed 180 days in 2018.

Step-by-step explanation:

Let's find out how many days in total Guido stayed in US in these 3 years 2017, 2018 and 2019.

Year = 2017

Days = 200

Year = 2018

Days = 180

Year = 2019

Days = 70

Total days stayed = 200 + 180 + 70

Total Days Stayed = 450 days.

In U.S, there are two tests are in place and for non-citizen of U.S and for the resident Alien or non - resident Alien status, one must pass one of these two tests, which are as follows:

1. Green Card Test:

2. Substantial Presence Test:

Here, in this problem, Guido is citizen and resident of Belgium. So, will check his criteria according to the substantial presence test.

So, the question is: how many days Guido was present in the U.S in 2018 under resident alien status.

In 2018, Guido stayed in US for 180 days.

So, according to the Substantial Presence test, one must be physically present in US for more than 31 days to be eligible for resident alien status. In addition, in 2018, his total of physical presence in US in 2018 and one third of physical presence in 2017 must be greater than 183 days.

If we see, both conditions are matched in case of Guido.

Guido stayed in US in 2018 for 180 days which are greater than 31 days.

Guido Stayed in US in 2018 and in 2017 = (180 + 66) > 183 days.

So, yes Guido does meet US Statutory definition in 2018 and stayed 180 days in 2018.

3 0

3 years ago

the number of mosquito at the beginning of the summer was 4,000. the population of mosquito is expected to grow at a rate of 25%

lara31 [8.8K]

Answer:

about 9765 mosquitoes

Step-by-step explanation:

This is an exponential growth problem which follows the formula:

$A=P(1+r)^t$

Where A is the future amount (we need this after 4 months)

P is the initial amount (here, it is 4000)

r is the rate of increase (here it is 25% or 0.25)

t is the time period (here t= 4, because we want to know after 4 months)

plugging in all the available/given info, we can solve for A:

$A=P(1+r)^t\\A=4000(1+0.25)^4\\A=4000(1.25)^4\\A=9765.63$

Since, no fractional mosquitoes, we round down to 9765 mosquitoes

3 0

3 years ago

Examples of lines geometry

brilliants [131]

Hope that’s what you were asking for:)

8 0

4 years ago

a person leaves their house to go to the store. it takes 15 minutes at 12 mph to get to the store. They return home at 31 mph. W

trasher [3.6K]

$\bf \cfrac{\stackrel{speed~forth}{12}+\stackrel{speed~back}{31}}{\stackrel{interval~forth}{1}+\stackrel{interval~back}{1}}\implies \cfrac{43}{2}\implies 21.5mph\textit{ on average}$

5 0

4 years ago

Which system of linear inequalities has the point (3, –2) in its solution set? y less-than negative 3. y less-than-or-equal-to t

xxTIMURxx [149]

Answer:

$y > -3$ and $y \geq \frac{2}{3}x- 4$

Step-by-step explanation:

The complete question is

Which system of linear inequalities has the point (3,-2) in its solution set?

A.

y < -3

y ≤ 2/3x - 4

B.

y > -3

y ≥ 2/3x - 4

C.

y < -3

y ≥ 2/3x - 4

D.

y > -2

y ≤ 2/3x - 4

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each case

Case A) we have

$y < -3$ ----> inequality A

$y \leq \frac{2}{3}x- 4$ ----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

$-2< -3$ ----> is not true

therefore

The ordered pair is not a solution of the system A

Case B) we have

$y > -3$ ----> inequality A

$y \geq \frac{2}{3}x- 4$ ----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

$-2< -3$ ----> is true

Inequality B

$-2 \geq \frac{2}{3} (3)-4$

$-2\geq -2$ ----> is true

therefore

The ordered pair is a solution of the system B

Case C) we have

$y< -3$ ----> inequality A

$y \geq \frac{2}{3}x- 4$ ----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

$-2< -3$ ----> is not true

therefore

The ordered pair is not a solution of the system C

Case D) we have

y > -2 ----> inequality A

$y \leq \frac{2}{3}x- 4$ ----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

$-2> -2$ ----> is not true

therefore

The ordered pair is not a solution of the system D

9 0

3 years ago