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myrzilka [38]
2 years ago
8

A. Draw a flowchart or write pseudocode to represent the logic of a program that allows the user to enter an hourly pay rate and

hours worked. The program outputs the user’s gross pay.
b. Modify the program that computes gross pay to allow the user to enter the withholding tax rate. The program outputs the net pay after taxes have been withheld
Computers and Technology
1 answer:
konstantin123 [22]2 years ago
6 0

Answer:

Here is the pseudocode.

a.

INPUT hourly rate

INPUT hours worked

SET gross pay = hourly rate  x hours worked

PRINT gross pay

b.

INPUT hourly rate

INPUT hours worked

SET gross pay = hourly rate  x hours worked

PRINT gross pay

INPUT tax rate

SET net pay = gross pay - (gross pay * tax rate / 100)

PRINT net pay

Explanation:

a.

Ask the user to enter hourly rate and hours worked

Calculate the gross pay, multiply hourly rate by hours worked

Print the gross pay

b.

Ask the user to enter hourly rate and hours worked

Calculate the gross pay, multiply hourly rate by hours worked

Print the gross pay

Ask the user to enter the tax rate

Calculate the net pay as gross pay - (gross pay * tax rate / 100)

Print the net pay

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July 9, 2015

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A search will _____ determine the index of the goal.
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The correct answer is C. Sometimes.

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Express the following binary numbers in hexadecimal. (a) %100011100101 (b) %1011001111 (show work)
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Answer:

(100011100101)_{2} = (8E5)_{16} = %8E5

(1011001111) = (2CF)_{16} = %2CF

Explanation:

Binary and hexadecimal values have the following pair equivalences.

(0000)_{2} = (0)_{16}

(0001)_{2} = (1)_{16}

(0010)_{2} = (2)_{16}

(0011)_{2} = (3)_{16}

(0100)_{2} = (4)_{16}

(0101)_{2} = (5)_{16}

(0110)_{2} = (6)_{16}

(0111)_{2} = (7)_{16}

(1000)_{2} = (8)_{16}

(1001)_{2} = (9)_{16}

(1010)_{2} = (A)_{16}

(1011)_{2} = (B)_{16}

(1100)_{2} = (C)_{16}

(1101)_{2} = (D)_{16}

(1110)_{2} = (E)_{16}

(1111)_{2} = (F)_{16}

We convert from binary to hexadecimal selecting groups of 4 binary from the binary code, from the least significant bits(at the right) to the most significant bits(at the left). The conversion is an hexadecimal "string" from the last group you converted to the first. So:

(a) %100011100101

(0101)_{2} = (5)_{16}

(1110)_{2} = (E)_{16}

(1000)_{2} = (8)_{16}

So

(100011100101)_{2} = (8E5)_{16}

(b) %1011001111

(1111)_{2} = F_{16}

(1100)_{2} = C_{16}

(10)_{2} = (0010)_{2} = 2_{16}

(1011001111) = (2CF)_{16}

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