Please answer this for me and show me the work.

Jason is considering two different jobs that he has been offered and he likes the atmosphere and

aniked [119]

The amount of money that Jason will receive in one year after all deductions have been taken for each company is $30,780.5 and $32562.

<h3>What is the final amount?</h3>

With the first job, he would be an apartment property manager for AMLI Company.

This job is offering Jason an hourly rate of $19. 00 an hour plus a weekly bonus of $40 for each new tenant he sets a one-year lease-up with.

The company expects Jason to work all 52 weeks of the year and 40 hours per week.

He was told he could expect to set up an average of 5 lease agreements for each week that he works.

The amount after one year will be given by an equation

Let y be the amount, x be the number of hours in a week and z be the number of the week.

y = 19x + 40z

The total number of hours in a year will be

x = 2080

Then

y = 19 × 2080 + 40 × 52

y = $ 41,600

The amount after the deduction of Social Security (6. 2%), Medicare (1. 2%), State Income Tax (3%), and Federal Income Tax (14%) will be

→ 41,600 × 0.938 × 0.988 × 0.97 × 0.86

→ $ 32,160.5

The amount after the insurance will be

Amount = 32160.5 - 115 × 12

Amount = $ 30780.5

With the second job, he would work as a water treatment facility manager in Gwinnett County. He would be paid an annual salary of $46,000 a year, there are no bonuses and he gets a week off.

The amount after the deduction of Social Security (6. 2%), Medicare (1. 2%), State Income Tax (3%), and Federal Income Tax (14%) will be

→ 46,000 × 0.938 × 0.988 × 0.97 × 0.86

→ $ 35,562

The amount after the insurance will be

Let the number of the worker be 10. Then

Amount = 35,562 - 10 × 12 × 25

Amount = 35562 - 3000

Amount = $32562

More about the final amount link is given below.

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5 0

1 year ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago

The graph shows the best-fit regression model for a set of data comparing the number of hours spent hiking and the number of mil

Naddik [55]

Answer:

C

Step-by-step explanation:

Edgen

3 0

2 years ago

Read 2 more answers

G(x) = –(x + 3)2 – 4?

professor190 [17]

Answer:

g(x)=-x-5

Step-by-step explanation:

Simplify parenthesis by distribution:

g(x)=-x-3+2-4

Addition and Subtraction:

-3+2=-1, -1-4=-5

Finalize since no more simplification:

g(x)=-x-5

Pls give thx and brainliest wanna level up

6 0

2 years ago

Let A be a set of numbers.

LUCKY_DIMON [66]

Answer:

a. closed under addition and multiplication

b. not closed under addition but closed under multiplication.

c. not closed under addition and multiplication

d. closed under addition and multiplication

e. not closed under addition but closed under multiplication

Step-by-step explanation:

a.

Let A be a set of all integers divisible by 5.

Let $x,y$ ∈A such that $x=5m\,,\,y=5n$

Find $x+y,xy$

$x+y=5m+5n=5(m+n)$

So, $x+y$ is divisible by 5.

$xy=(5m)(5n)=25mn=5(5mn)$

So,

$xy$ is divisible by 5.

Therefore, A is closed under addition and multiplication.

b.

Let A = { 2n +1 | n ∈ Z}

Let $x,y$ ∈A such that $x=2m+1\,,\,y=2n+1$ where m, n ∈ Z.

Find $x+y,xy$

$x+y=2m+1+2n+1=2m+2n+2=2(m+n+1)$

So,

$x+y$ ∉ A

$xy=(2m+1)(2n+1)=4mn+2m+2n+1=2(2mn+m+n)+1$

So,

$xy$ ∈ A

Therefore, A is not closed under addition but A is closed under multiplication.

c.

$Let A=\{2,5,8,11,14,...\}$

Let $x=2,y=5$ but $x+y=2+5=7$ ∉A

Also,

$xy=2(5)=10$ ∉A

Therefore, A is not closed under addition and multiplication.

d.

Let A = { 17n: n∈Z}

Let $x,y$ ∈ A such that $x=17n,y=17m$

Find x + y and xy

$x+y=17n+17m=17(n+m)$

$xy=(17m)(17n)=289mn=17(17mn)$

So,

$x+y,xy$ ∈ A

Therefore, A is closed under addition and multiplication.

e.

Let A be the set of nonzero real numbers.

Let $x,y$ ∈ A such that $x=-1,y=1$

Find x + y

$x+y=-1+1=0$

So,

$x+y$ ∈ A

Also, if x and y are two nonzero real numbers then xy is also a non-zero real number.

Therefore, A is not closed under addition but A is closed under multiplication.

4 0

2 years ago