4. Acme Inc. has 200 total employees, 150 of which are nonexcludable employees. Ten employees are highly compensated. Seven of the 10 highly compensated and 100 of the 140 nonhighly compensated employees are covered under Acme’s qualified plan. The average accrued benefits for the highly compensated is 3% and the average accrued benefit for the nonhighly compensated is 1.5%. Which of the following statements is true regarding coverage?
1. The plan passes the ratio percentage test.
2. The plan passes the average benefits test.
1. 1 only.
2. 2 only.
3. Both 1 and 2.
4. Neither 1 nor 2.
In: Accounting
Calculate Mean, Median, Mode, Quartiles, Percentiles,
Population Variance, and Standard deviation from the
following grouped data:
Class.................. Frequency
2 - 4 ................. 3
4 - 6 ..................... 4
6 - 8 .................... 2
8 - 10 ................... 1
In: Statistics and Probability
3.Given is a Decision Tree Diagram. The Payoffs 1-14 are given
in the table below. Answer questions a, b, and c.
|
Payoff |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
|
$ |
6 |
-3 |
5 |
5 |
8 |
5 |
5 |
-1 |
5 |
4 |
5 |
2 |
7 |
5 |
a) The value at node 4 is
b) The value at node 8 is (in 1 decimal place)
c) The best course of action or decision is to select
alternative
In: Statistics and Probability
4(a)Verify that the equation y=lnx-lny is an implicit solution of the IVP,
ydx=x(y+1)dy
y(e)=1
(b) Consider the IVP, test whether it is exact and solve it.
x(x-y)dy=-(3xy-y^2)dx
y(-1)=1
(c) Determine the degree of the following homogeneous function
(i) f(x,y)=4x^2+2y^4 ,
(ii) f(x,y)=√5x^6-3y^6+4x^2y^4 ,
In: Advanced Math
Let Q1, Q2, Q3 be constants so that (Q1, Q2) is the critical point of the function f(x, y) = (175)x 2 + (−150)xy + (175)y 2 + (−200)x + (400)y + (230), and Q3 = 1 if f has a local minimum at (Q1, Q2), Q3 = 2 if f has a local maximum at (Q1, Q2), Q3 = 3 if f has a saddle point at (Q1, Q2), and Q3 = 4 otherwise. Let Q = ln(3 + |Q1| + 2|Q2| + 3|Q3|). Then T = 5 sin2 (100Q)
satisfies:— (A) 0 ≤ T < 1. — (B) 1 ≤ T < 2. — (C) 2 ≤ T < 3. — (D) 3 ≤ T < 4. — (E) 4 ≤ T ≤ 5
In: Physics
Let Q1, Q2, Q3 be constants so that (Q1, Q2) is the critical point of the function f(x, y) = (90)x 2 + (0)xy + (90)y 2 + (−72)x + (96)y + (40), and Q3 = 1 if f has a local minimum at (Q1, Q2), Q3 = 2 if f has a local maximum at (Q1, Q2), Q3 = 3 if f has a saddle point at (Q1, Q2), and Q3 = 4 otherwise. Let Q = ln(3 + |Q1| + 2|Q2| + 3|Q3|). Then T = 5 sin2 (100Q)
satisfies:— (A) 0 ≤ T < 1. — (B) 1 ≤ T < 2. — (C) 2 ≤ T < 3. — (D) 3 ≤ T < 4. — (E) 4 ≤ T ≤ 5
In: Advanced Math
1. What is a divestiture? (2)
2. Briefly explain the difference between a spin-off and a carve-out. (4)
3. Briefly explain bankruptcy costs(4)
In: Economics
Using repeated measure on SPSS - I need tests of within and pairwise comparisons
( Upload screenshot of SPSS - tests of within and pairwise comparisons )
A group of 8 second-grade students receive 3 different kinds of reinforcement for responding to teacher questioning, no reinforcement, verbal reinforcement, and a tangible reinforcer (a sticker) for responding. The dependent variable is the number of respondents.
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Non |
Verb |
Token |
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1 |
2 |
4 |
5 |
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2 |
3 |
5 |
6 |
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3 |
3 |
4 |
7 |
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4 |
4 |
6 |
7 |
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5 |
6 |
6 |
8 |
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6 |
2 |
4 |
5 |
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7 |
1 |
3 |
4 |
|
8 |
2 |
5 |
7 |
In: Statistics and Probability
Program 5: Decimal to Binary
In Computer Science we often work with different number systems to represent data. The most commonly used number system (which we refer to as "decimal") is a "base 10" number system, which means that we use the values 0-9 to represent all numbers.
The binary number system is a "base 2" number system, which means that we use the values 0 and 1 to represent all numbers. We refer to each value in a binary number as a "bit" (i.e. the number 0101 consists of 4 bits). Here are some numbers represented in both base 10 and base 2:
| Decimal | Binary |
| 0 | 0 |
| 1 | 1 |
| 2 | 10 |
| 3 | 11 |
| 10 | 1010 |
| 25 | 11001 |
| 65 | 1000001 |
| 127 | 1111111 |
| 128 | 10000000 |
This is the algorithm for converting a decimal number into a binary number:
Note that each successive bit you compute is for the next highest place (the next highest power of 2) -- so it will be inserted at the front (or the left-end) of the binary number.
For example, consider the following decimal number:
decimal = 5
Here is how the algorithm works on this number:
5 ÷ 2 = 2 remainder 1 Record this bit: 1
5 // 2 = 2
2 ÷ 2 = 1 remainder 0 Record this bit: 01
2 // 2 = 1
1 ÷ 2 = 0 remainder 1 Record this bit: 101
1 // 2 = 0 (algorithm stops)
For this problem, write a program that prompts the user for an integer greater than or equal to zero. If the user supplies a value less than 0 you should re-prompt them. Then convert the number to binary, printing out each step in the process as you go.
Note: Your binary value can't be recorded as an integer. Since you are building the number up bit-by-bit, you will probably need to keep track of 0's that are in higher place values than the the highest 1 bit, in other words, leading zeros. Integers do not support leading 0's (i.e. 0001 will be reduced to 1). Therefore, for this program you should use a String to hold your bits (i.e. "0001" is allowed). You will need to figure out a way how to add bits to your String (hint: concatenation)
Here are a few sample runnings of your program:
Enter a number greater than or equal to 0: -5 Invalid, try again Enter a number greater than or equal to 0: -9 Invalid, try again Enter a number greater than or equal to 0: 9 9 % 2 = 1 ---> 1 9 // 2 = 4 4 % 2 = 0 ---> 01 4 // 2 = 2 2 % 2 = 0 ---> 001 2 // 2 = 1 1 % 2 = 1 ---> 1001 1 // 2 = 0 9 in binary is 1001
Enter a number greater than or equal to 0: 86 86 % 2 = 0 ---> 0 86 // 2 = 43 43 % 2 = 1 ---> 10 43 // 2 = 21 21 % 2 = 1 ---> 110 21 // 2 = 10 10 % 2 = 0 ---> 0110 10 // 2 = 5 5 % 2 = 1 ---> 10110 5 // 2 = 2 2 % 2 = 0 ---> 010110 2 // 2 = 1 1 % 2 = 1 ---> 1010110 1 // 2 = 0 86 in binary is 1010110
Enter a number greater than or equal to 0: 255 255 % 2 = 1 ---> 1 255 // 2 = 127 127 % 2 = 1 ---> 11 127 // 2 = 63 63 % 2 = 1 ---> 111 63 // 2 = 31 31 % 2 = 1 ---> 1111 31 // 2 = 15 15 % 2 = 1 ---> 11111 15 // 2 = 7 7 % 2 = 1 ---> 111111 7 // 2 = 3 3 % 2 = 1 ---> 1111111 3 // 2 = 1 1 % 2 = 1 ---> 11111111 1 // 2 = 0 255 in binary is 11111111
Enter a number greater than or equal to 0: 256 256 % 2 = 0 ---> 0 256 // 2 = 128 128 % 2 = 0 ---> 00 128 // 2 = 64 64 % 2 = 0 ---> 000 64 // 2 = 32 32 % 2 = 0 ---> 0000 32 // 2 = 16 16 % 2 = 0 ---> 00000 16 // 2 = 8 8 % 2 = 0 ---> 000000 8 // 2 = 4 4 % 2 = 0 ---> 0000000 4 // 2 = 2 2 % 2 = 0 ---> 00000000 2 // 2 = 1 1 % 2 = 1 ---> 100000000 1 // 2 = 0 256 in binary is 100000000
This program should be named as follows: LastNameFirstName_assign4_problem5.py (for example, "KappCraig_assign4_problem5.py")
In: Computer Science
C++ Use BinaryNodeTree to solve the following questions in a program:
1) What is the value of the prefix expression +−∗ 235/↑ 2 3 4?
2) What is the postfix form of the expression ((x + y) ↑ 2) + ((x − 4)/3)?
3) What is the value of the postfix expression723 ∗ − 4 ↑ 9 3/+?
In: Computer Science