ELA 5 Summit C: 78%

Early American Art Summit
C: 72%

Early American History 5 Summit MI
B: 81%

Math 5 Summit
D: 69%

Online Learning K-8 19-20
C: 80%

PE 5 Summit
A: 97%

Science 5 Summit NG
C: 76%

Stride ELA 5
A: 99%
Stride Math 5
A: 99%
Stride Science 5
A: 100%

will I have to retake my grade?? (END OF SCHOOL JUNE 16TH) Also I can always go back and do my work, its easy to say ill do it, but its harder when I have stuff to do... ;c

In outlook how can cherie look at the details of an event that appears on the month view of her calendar? check all that apply. by switching to the detail view by switching to the week view by switching to the day view by right-clicking on the event by double-clicking on the event by highlighting the event

Assume the existence of a bankaccount class. define a derived class, savingsaccount that contains two instance variables: the first a double, named interestrate, and the second an integer named interesttype. the value of the interesttype variable can be 1 for simple interest and 2 for compound interest. there is also a constructor that accepts two parameters: a double that is used to initialize the interestrate variable, and a string that you may assume will contain either "simple", or "compound", and which should be used to initialize the interesttype variable appropriately. there should also be a pair of functions getinterestrate and getinteresttype that return the values of the corresponding data members (as double and int respectively).

Consider the following statements: struct nametype{string first; string last; }; struct coursetype{string name; int callnum; int credits; char grade; }; struct studenttype{nametype name; double gpa; coursetype course; }; studenttype student; studenttype classlist[100]; coursetype course; nametype name; mark the following statements as valid or invalid. if a statement is invalid, explain why.a.) student.course.callnum = "csc230"; b.) cin > > student.name; c.) classlist[0] = name; d.) classlist[1].gpa = 3.45; e.) name = classlist[15].name; f.) student.name = name; g.) cout < < classlist[10] < < endl; h.) for (int j = 0; j < 100; j++)classlist[j].name = name; i.) classlist.course.credits = 3; j.) course = studenttype.course;