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BARSIC [14]
2 years ago
3

Enzo’s priority is to build more skateboards and get more customers, but his company has been losing customers due to the amount

of time it takes to build a custom skateboard. He needs to focus on time-effectiveness over cost-efficiency for a month to be able to acquire more customers. What should Enzo do? Hire two more employees and spend an additional $3,000 on compensation to produce 1,200 skateboards for the month. Hire one more employee and spend $1,500 on compensation to produce 600 skateboards for the month. Get help from a friend at no cost to produce 500 skateboards for the month. Work day and night to produce 400 skateboards for the month. LIMITED TIME HELP!!
Computers and Technology
2 answers:
Natalka [10]2 years ago
7 0

Answer:

its answer a

Explanation:

i just took the quiz

ladessa [460]2 years ago
4 0

Answer:

A: Hire two more employees and spend an additional $3,000 on compensation to produce 1,200 skateboards for the month.

Explanation:

just did the test :)

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Secure Wi-Fi networks and VPNs use _____ to secure data transferred over a network.
Aleksandr-060686 [28]

VPNs and Wifi networks use tunneling to send data privately over a network

4 0
2 years ago
Problem 1a. Write a function named hasFinalLetter that takes two parameters 1. strList, a list of non-empty strings 2. letters,
mash [69]

Answer:

The answer is the programming in Python language has strings and characters that has to be declared in the method.

Explanation:

#method

def hasFinalLetter(strList,letters):

output = #output list

#for every string in the strList

for string in strList:

#findout the length of each string in strList

length = len(string)

#endLetter is last letter in each string

endLetter = string[length-1]

#for each letter in the letters list

for letter in letters:

#compare with endLetter

#if we found any such string

#add it to output list

if(letter == endLetter):

output.append(string)

#return the output list

return output

#TestCase 1 that will lead to return empty list

strList1 = ["user","expert","login","compile","Execute","stock"]

letters1 = ["a","b","y"]

print hasFinalLetter(strList1,letters1)

#TestCse2

strList2 = ["user","expert","login","compile","Execute","stock"]

letters2 = ["g","t","y"]

print hasFinalLetter(strList2,letters2)

#TestCase3

strList3 = ["user","expert","login","compile","Execute","stock"]

letters3 = ["k","e","n","t"]

print hasFinalLetter(strList3,letters3)

8 0
2 years ago
Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
Miles to track laps One lap around a standard high-school running track is exactly 0.25 miles. Write a program that takes a numb
fredd [130]

Answer:

Explanation:

#include <iostream>

using namespace std;

//This is the function in the next line

double MilesToLaps(double userMiles) {

double Laps;  

Laps = userMiles * 4;

return Laps

}

//This is the main fucnction

int main(){

double Miles, FinLap;

cout << "Enter the number of miles " << endl;

cin>>Miles;

FinLap = MilesToLaps(Miles);

cout << "The number of laps ran is: "<<setprecision(2)<<Finlap<<endI;

}

3 0
2 years ago
Read 2 more answers
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Aleksandr-060686 [28]

Answer:

RAID 1

Explanation:

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RAID 1

Disk mirroring, also known as RAID 1, is the replication of data to two or more disks. Disk mirroring is a good choice for applications that require high performance and high availability, such as transnational applications, email and operating systems. Disk mirroring also works with solid state drives so “drive monitoring” may be a better term for contemporary storage systems.

8 0
2 years ago
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