1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
slamgirl [31]
3 years ago
6

A paycheck is issued for $329.40. The paystub reflects an amount earned of $400.00, Medicare tax of $5.80, Social Security tax o

f $24.80, and a federal tax of $40.00. What is the net income for the paycheck? Group of answer choices
Mathematics
2 answers:
jok3333 [9.3K]3 years ago
5 0

Answer:

gross income is $400

The net is $329.40

Deductions are $$70.6

Step-by-step explanation:

Goryan [66]3 years ago
4 0

Answer:

$329.40

Step-by-step explanation:

A pay stub is a document that contains the pay, salary or wages of an employee.

A pay stub contains the following information:

a) Net pay of an employee

b) Tax deductions, e.t.c

The net pay or income of an employee is calculated after subtracting all deductions from their Gross pay.

From the above question we have the following information:

Amount earned or Gross pay = $400.00

Medicare tax = $5.80,

Social Security tax = $24.80

Federal tax = $40.00

Net Income =

= $400 - $(24.80 + 5.80 + 40)

= $400 - $70.60

= $329.40

The net income for the paycheck = $329.40

You might be interested in
F(x)=-2x+1; {-2,0,2,4,6} Find the range of the function for the given domain.
3241004551 [841]

Answer:

For the domain indicated, the range is  {-11, -7, -3, 1, 5}

Step-by-step explanation:

Given the indicated domain of f(x) you have to replace the values in the function to find the range (the "y" values of the function).

f(-2) = 5

f(0) = 1

f(2) = -3

f(4) = -7

f(6) = -11.

So the range for this domain is {-11, -7, -3, 1, 5}

4 0
3 years ago
Which statement describes this system of equations? 9x – 6y = 15, 3x – 2y = 5 The equations in the system are equivalent equatio
damaskus [11]

Answer:

There is no solution to the systems of equation.

Step-by-step explanation:

Graph the system by using y=mx+b

Both systems are y=2/5x+5/2.

8 0
3 years ago
Read 2 more answers
The value of the pronumeral​
Sergio039 [100]

Answer: stop pay attention in class

Step-by-step explanation:

7 0
3 years ago
Evaluate 0.3 y+\dfrac yz0.3y+ z y ​ 0, point, 3, y, plus, start fraction, y, divided by, z, end fraction when y=10y=10y, equals,
sergij07 [2.7K]

Answer:

  5

Step-by-step explanation:

Put the numbers where the corresponding variables are in the formula and do the arithmetic.

  0.3y+\dfrac{y}{z}=0.3\cdot 10+\dfrac{10}{5}=3+2=5

The expression evaluates to 5.

6 0
3 years ago
The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f
GenaCL600 [577]

Answer:

Concave Up Interval: (- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)

Concave Down Interval: (\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )

General Formulas and Concepts:

<u>Calculus</u>

Derivative of a Constant is 0.

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Second Derivative Test:

  • Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefined
  • Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
  • Number Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

<u>Step 1: Define</u>

f(x)=\frac{3}{1+x^2}

<u>Step 2: Find 2nd Derivative</u>

  1. 1st Derivative [Quotient/Chain/Basic]:                           f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}
  2. Simplify 1st Derivative:                                                           f'(x)=\frac{-6x}{(1+x^2)^2}
  3. 2nd Derivative [Quotient/Chain/Basic]:     f"(x)=\frac{-6(1+x^2)^2-2(1+x^2) \cdot 2x \cdot -6x}{((1+x^2)^2)^2}
  4. Simplify 2nd Derivative:                                                       f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}

<u>Step 3: Find P.P.I</u>

  • Set f"(x) equal to zero:                    0=\frac{6(3x^2-1)}{(1+x^2)^3}

<em>Case 1: f" is 0</em>

  1. Solve Numerator:                           0=6(3x^2-1)
  2. Divide 6:                                          0=3x^2-1
  3. Add 1:                                              1=3x^2
  4. Divide 3:                                         \frac{1}{3} =x^2
  5. Square root:                                   \pm \sqrt{\frac{1}{3}} =x
  6. Simplify:                                          \pm \frac{\sqrt{3}}{3}  =x
  7. Rewrite:                                          x= \pm \frac{\sqrt{3}}{3}

<em>Case 2: f" is undefined</em>

  1. Solve Denominator:                    0=(1+x^2)^3
  2. Cube root:                                   0=1+x^2
  3. Subtract 1:                                    -1=x^2

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is x= \pm \frac{\sqrt{3}}{3} (x ≈ ±0.57735).

<u>Step 4: Number Line Test</u>

<em>See Attachment.</em>

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

  1. Substitute:                    f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(1)-1)}{(1+1)^3}
  3. Multiply:                        f"(x)=\frac{6(3-1)}{(1+1)^3}
  4. Subtract/Add:              f"(x)=\frac{6(2)}{(2)^3}
  5. Exponents:                  f"(x)=\frac{6(2)}{8}
  6. Multiply:                       f"(x)=\frac{12}{8}
  7. Simplify:                       f"(x)=\frac{3}{2}

This means that the graph f(x) is concave up before x=\frac{-\sqrt{3}}{3}.

x = 0

  1. Substitute:                    f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(0)-1)}{(1+0)^3}
  3. Multiply:                       f"(x)=\frac{6(0-1)}{(1+0)^3}
  4. Subtract/Add:              f"(x)=\frac{6(-1)}{(1)^3}
  5. Exponents:                  f"(x)=\frac{6(-1)}{1}
  6. Multiply:                       f"(x)=\frac{-6}{1}
  7. Divide:                         f"(x)=-6

This means that the graph f(x) is concave down between  and .

x = 1

  1. Substitute:                    f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(1)-1)}{(1+1)^3}
  3. Multiply:                       f"(x)=\frac{6(3-1)}{(1+1)^3}
  4. Subtract/Add:              f"(x)=\frac{6(2)}{(2)^3}
  5. Exponents:                  f"(x)=\frac{6(2)}{8}
  6. Multiply:                       f"(x)=\frac{12}{8}
  7. Simplify:                       f"(x)=\frac{3}{2}

This means that the graph f(x) is concave up after x=\frac{\sqrt{3}}{3}.

<u>Step 5: Identify</u>

Since f"(x) changes concavity from positive to negative at x=\frac{-\sqrt{3}}{3} and changes from negative to positive at x=\frac{\sqrt{3}}{3}, then we know that the P.P.I's x= \pm \frac{\sqrt{3}}{3} are actually P.I's.

Let's find what actual <em>point </em>on f(x) when the concavity changes.

x=\frac{-\sqrt{3}}{3}

  1. Substitute in P.I into f(x):                    f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}
  2. Evaluate Exponents:                          f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }
  3. Add:                                                    f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }
  4. Divide:                                                f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}

x=\frac{\sqrt{3}}{3}

  1. Substitute in P.I into f(x):                    f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}
  2. Evaluate Exponents:                          f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }
  3. Add:                                                    f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }
  4. Divide:                                                f(\frac{\sqrt{3}}{3} )=\frac{9}{4}

<u>Step 6: Define Intervals</u>

We know that <em>before </em>f(x) reaches x=\frac{-\sqrt{3}}{3}, the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that <em>after </em>f(x) passes x=\frac{\sqrt{3}}{3}, the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: (- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)

We know that <em>after</em> f(x) <em>passes</em> x=\frac{-\sqrt{3}}{3} , the graph is concave up <em>until</em> x=\frac{\sqrt{3}}{3}. We used the 2nd Derivative Test to confirm this.

Concave Down Interval: (\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )

6 0
2 years ago
Other questions:
  • 4. Evaluate the geometric series given a1 = 4, r = -2 and n = 12
    12·1 answer
  • The point (2, 5) is represented on the graph. Which point also belongs on this graph?
    12·2 answers
  • Determine if the proportion is true
    6·2 answers
  • 2. <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B9%7D%20" id="TexFormula1" title=" \sqrt{9} " alt=" \sqrt{9} " align="absmi
    15·2 answers
  • What is 0.689 in word form
    9·1 answer
  • Help! step by step please!
    11·1 answer
  • Alguien me pude decir una adivinanza o historia q la solución sea una ecuación de segundo grado completa????
    14·1 answer
  • Given m || n, find the value of x and y.<br> m(7x+13)<br> n(3y-17) (9x-19)
    6·1 answer
  • Hi can someone explain what does consecutive angles are supplementary mean
    5·1 answer
  • Dawn is verifying the accuracy of her paycheck. She
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!