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Given that Jamie worked 50 hours (Hours = 50) last week and earns $10.00 an hour (Rate = 10), how much did Jamie earn last week,

quester [9]

Answer:

Jamie earned (Total pay) $500.

Explanation:

We are given the following code:

If (Rate >=10) OR (Hours <=40) Then

TotalPay = Hours * Rate

Else

TotalPay = (Hours * Rate)+(Hours–40)*Rate*1.5

End If

Let us understand the code line by line:

The first line contains an if statement with 2 conditions:

i.e. 1st condition:

The rate is greater than or equal to 10

2nd condition:

Number of hours are lesser than or equal to 40.

There is OR between the two condition i.e. the statement next to if() statement will get executed if any one of them becomes true and else part will not be executed.

The next statement is:

TotalPay = Hours * Rate

It calculates the pay if any of the two conditions written earlier becomes true.

Next statement is else statement:

It will get executed given that the above if() statement becomes false.

Now, we are given that Jamie worked 50 hours last week and earns $10.00 an hour:

i.e.

Hours = 50

Rate = 10

Now, let get to the code execution.

The first condition is true i.e. Rate >= 10 (because Rate is 10 here)

So, the following statement will be used to calculate the Total pay:

TotalPay = Hours * Rate

and else part will not be executed.

TotalPay = 50 * 10 = $500

Jamie earned (Total pay) $500.

3 0

3 years ago

To make a complicated task easier, use a _____.

tia_tia [17]

I believe the answer is B spread sheet because i use spread sheets to make my life easier. I organize info in a spread sheet then it is really easy to find.

hope this helps

6 0

3 years ago

At one college, the tuition for a full-time student is $6,000 per semester. It has been announced that the tuition will increase

Alina [70]

Answer:

var projected_fee = 6000;
for(var i = 1; i <= 5; i++){
projected_fee = projected_fee * 0.02 + projected_fee;
console.log("$" + projected_fee.toFixed(2));
}

Explanation:

Firstly, create a variable, projected_fee, and set the initial tuition fee value to it (Line 1).

Next, user a for loop that run for 5 times to repeatedly calculate the projected_fee based on 2 percent of increment rate (Line 4) and display the projected fee to console terminal (Line 5). The output should be

$6120.00

$6242.40

$6367.25

$6494.59

$6624.48

8 0

3 years ago

What is a quasi vpn?

Natasha2012 [34]

Answer:

i don't think there is such thing as a quasi vpn

Explanation:

6 0

3 years ago

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

2 years ago