A game has 50 cards with 10 each numbered 1 to 5,and a player must draw a 2 or a 3 to move out of the “start” position.

Design minimal 2-level AND-OR and NAND-NAND realizations of the following logic function. Draw circuit diagrams of your realizat

Delicious77 [7]

Answer:

Realization :

since there are 3-variable. Therefore 2ⁿ = 2³ = 8

From the attached solution, f(x,y,z) = y' + xz'

also' see the circuit diagrams of my realizations of AND-OR and NAND-NAND

Step-by-step explanation:

See attached picture for minimal 2-level AND-OR and NAND-NAND of the logic function.

f(x,y,z) = y' + xz' was designed from the function f(x,y,z) = Σm(0,1,4,5,6)

Then, minimal 2-level AND-OR was designed and after was NAND-NAND designed as well.

5 0

3 years ago

you invest $2800 in an account that pays an interest of 5.5%, compounded continuously. calculate the balance of your account aft

LuckyWell [14K]

Hi, hope this helps you. Have a good day. :)

Answer:

A = $5,417.42 (balance of the account)

I = $2,617.42

5 0

3 years ago

Read 2 more answers

Greg have 24 more stickers than Wayne and Ronald have twice as much stickers than Wayne together they have 124 stick how much ea

lukranit [14]

G=24+w
r=2w
g+w+r=124
Substitute the values to get 24+w+w+2w=124
Simplifies to 4w=100. Therefore, Wayne has 25. Greg has 24 more than Wayne, so he has 49. Ronald would have 50.

4 0

3 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

a

$P(X \ge 1) = 0.509$

b

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$