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ohaa [14]
2 years ago
10

3D-bioprinted human organs may be subject to ________. Select one: a. Conflicting healthcare regulations b. Conflicting religiou

s, political, moral, and financial interests c. Skills of bio-engineers d. Technology limitations
Computers and Technology
1 answer:
tatuchka [14]2 years ago
4 0

Answer:

a. Conflicting healthcare regulations

Explanation:

3D-bioprinted human organs may be subject to  a. Conflicting healthcare regulations

The omnipresent essence of health-care legislation stems from the basic issues at stake. Most observers recognize the need for some form of supervision when factors as important as life and health are involved. Also those who are especially skeptical of heavy-handed bureaucracy in government see public interest in some sort of external oversight of this field

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Type the correct answer in the box. Spell all words correctly.
wlad13 [49]

Ryan should apply a filter/criteria on the Courses column and view all the courses that show Sociology

3 0
2 years ago
The attack in which the attacker sends a forged packet with the same source IP address and destination IP address in which the v
Sergeeva-Olga [200]

Answer:

TEARDROP

Explanation:

Teardrop is a form of attack in which the attacker sends a forged packet with the same source IP address and destination IP address in which the victim may be tricked into sending messages to and from itself .

Teardrop attack also involves sending fragmented packets to a target machine in which the victim is been tricked into sending messages to and from itself. One of the fields in an IP header is the “fragment offset” field, indicating the starting position of the data contained in a fragmented packet relative to the data in the original packet.

7 0
2 years ago
Taylor and Rory are hosting a party. They sent out invitations, and each one collected responses into dictionaries, with names o
disa [49]

Answer:

Following are the code to this question:

def combine_guest(guest1, guest2):#defining a method combine_guest that accepts two dictionary

   guest2.update (guest1)#use dictionary guest2 that use update method to update guest2 dictionary

   return guest2#return guest2 dictionary values

Rory_guest= { "Ada":2, "Ben":3, "Dav":1, "Jo":3, "Charry":2, "Terry":1, "bob":4}#defining a dictionary and add value

Taylor_guest = { "Dav":4, "Nan":1, "bobert":2, "Ada":1, "Samantha":3, "Chr":5}#defining a dictionary and add value

print(combine_guest(Rory_guest,Taylor_guest))#calling the combine_guest method

Output:

{'Nan': 1, 'Samantha': 3, 'Ada': 2, 'bob': 4, 'Terry': 1, 'Jo': 3, 'Ben': 3, 'Dav': 1, 'Charry': 2, 'bobert': 2, 'Chr': 5}

Explanation:

In the code a method, "combine_guest" is defined, that accepts two dictionaries "guest1, guest2" inside the method, in which the "guest2" dictionary uses the update method to combine the value of the guest1 dictionary and use a return keyword to return guest2 values.

In the next step, two dictionaries are declared, that holds some values and use a print method to call the "combine_guest" method and prints its return values.  

7 0
1 year ago
Given sphereRadius and piVal, compute the volume of a sphere and assign to sphereVolume. Look up the equation online (e.g., http
fenix001 [56]

Answer:

  1. #include <stdio.h>
  2. int main()
  3. {
  4.    const double piVal = 3.14159;
  5.    double sphereVolume = 0.0;
  6.    double sphereRadius = 0.0;
  7.    
  8.    sphereRadius = 1.0;
  9.    sphereVolume = 4.0/ 3.0 * piVal * sphereRadius * sphereRadius * sphereRadius;
  10.    
  11.    
  12.    printf("Sphere volume: %lf\n", sphereVolume);
  13.    return 0;
  14. }

Explanation:

Firstly we can identify the formula to calculate volume of sphere which is

Volume = 4/3 \pi r^{3}

With this formula in mind, we can apply this formula to calculate the volume of sphere in Line 10. This is important to perform floating-point division 4.0/3.0 to ensure the resulting value is a floating value as well. Since we have been given piVal and sphereRadius, we can just multiply the result of floating-point division with piVal and sphereRadius and get the sphereVolume value.

At last, display the sphere volume using printf method (Line 13).

6 0
2 years ago
Consider the following skeletal C program: void fun1(void); /* prototype */ void fun2(void); /* prototype */ void fun3(void); /*
natita [175]

Answer:

Check the explanation

Explanation:

a) main calls fun1; fun1 calls fun2; fun2 calls fun3

fun3()                                        d, e, f

fun2()                                        c, d, e

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "c" of fun2() will be visible

   variable "b" of fun1() will be visible

   variable "a" of main() will be visible

b) main calls fun1; fun1 calls fun3

fun3()                                        d, e, f

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "b, c" of fun1() will be visible

   variable "a" of main() will be visible

c) main calls fun2; fun2 calls fun3; fun3 calls fun1

fun1()                                        b, c, d

fun3()                                        d, e, f

fun2()                                        c, d, e

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun1().

   In fun1(), the local variables "b, c, d" of fun1() will be visible

   variable "e, f" of fun3() will be visible

   variable "a" of main() will be visible

d) main calls fun1; fun1 calls fun3; fun3 calls fun2

fun2()                                        c, d, e

fun3()                                        d, e, f

fun1()                                        b, c, d,

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun2().

   In fun2(), the local variables "c, d, e" of fun2() will be visible

   variable "f" of fun3() will be visible

     variable "b" of fun1() will be visible

   variable "a" of main() will be visible

The last function called will comprise of all its local variables and the variables other than its local variables from all its preceding function calls till the main function.

8 0
1 year ago
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