Please help!!! what are the 3 solutions of 0

Write the computer program for;urgent

AlladinOne [14]

ANSWER:

#include <stdio.h>

#define PAYRATE 10 //basic pay rate per hour

#define OVERTIME 15 //in excess of 40 hours a week

int NetPay(int hours);

int main()

{

int userWeeklyHours;

printf("Please enter your total weekly working hours: \n");

scanf("%d", &userWeeklyHours);// get the user weekly hours

NetPay(userWeeklyHours);

return 0;

}

int NetPay(int hours)// implementing the function to calculate total pay, total taxes, and net pay

{

int firstRate, secondRate, restOfRate, secondAmount, rest, payAfterTax, payedBeforeTax, overHours;

if (hours > 40)

{

overHours = hours - 40;

payedBeforeTax = (40 * PAYRATE) + (overHours * OVERTIME);

}

else

payedBeforeTax = hours * PAYRATE;//defining the user first paycheck before taxes

if (payedBeforeTax <= 300)//paying only first rate case

{

firstRate = payedBeforeTax*0.15;

payAfterTax =payedBeforeTax - firstRate;

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate, payAfterTax);

}

else if (payedBeforeTax > 300 && payedBeforeTax <= 450)//paying first and second rate

{

secondAmount = payedBeforeTax - 300;

firstRate = (payedBeforeTax - secondAmount) * 0.15;

secondRate = secondAmount * 0.20;

payAfterTax = payedBeforeTax - (firstRate + secondRate);

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate + secondRate, payAfterTax);

}

else if (payedBeforeTax > 450)// paying all rates

{

rest = payedBeforeTax - 450;

secondAmount = (payedBeforeTax - 300) - rest;

firstRate = (payedBeforeTax - (rest + secondAmount)) * 0.15;

secondRate = secondAmount * 0.20;

restOfRate = rest * 0.25;

payAfterTax = payedBeforeTax - (firstRate + secondRate + restOfRate);

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate + secondRate + restOfRate, payAfterTax);

}

return payAfterTax;

}

8 0

3 years ago

I need help with this math problem if u out a link I’ll report u but if u giv a good answer I’ll mark as brainliest

Sliva [168]

Answer:

1. MAD for Jada's data set = 1

Step-by-step explanation:

Mean Absolute Deviation = sum of the distances of each data value from the mean ÷ number of data value

Jada's data set is given as: 4, 4, 4, 6, 6 ,6

Mean = 5

✔️Find the distance between each data value form the mean

4 = |4 - 5| = 1

6 = |6 - 5| = 1

✔️Find the sum of the distances of each data value from the mean

1 + 1 + 1 + 1 + 1 + 1 = 6

✔️Find the MAD:

MAD = 6/6

MAD = 1

8 0

3 years ago

Solve for x<br> 5x + 6 < 3x + 7

Illusion [34]

Answer:

$5x + 6 < 3x + 7 \\ 5x - 3x < 7 - 6 \\ 2x < 1 \\ x < \frac{1}{2}$

8 0

3 years ago

Read 2 more answers

4 percent of x equals 56

brilliants [131]

X*.04 = 56
x = 56/.04
x = 1400
(* means "times")

7 0

3 years ago

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$