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3 years ago

Need help on Assignment 4: Evens and Odds

Computers and Technology

1 answer:

Sliva [168]3 years ago

8 0

n = int(input("How many numbers do you need to check? "))

even = 0

odd = 0

for x in range(n):

num = int(input("Enter number: "))

if num % 2 == 0:

even += 1

print(str(num) + " is an even number.")

else:

odd += 1

print(str(num) + " is an odd number.")

print("You entered " + str(even) + " even number(s).")

print("You entered " + str(odd) + " odd number(s).")

This works for me. Best of luck.

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3 years ago

Read 2 more answers

A network administrator is configuring an ACL with the command access-list 10 permit 172.16.32.0 0.0.15.255. Which IPv4 address

lord [1]

Answer:

For this wild card mask 0.0.15.255 the ACE IP address will be 172.16.47.254

Explanation:

ACL is the access control list that is used to enlist the ip addresses that allowed or restricted to access the network. ACE is an IP address from the list ACL that has all rules and regulations related to access of network. The ACE could be in the range of IP address in ACL. ACL can be calculated with the help of initial IP address adding with wild card mask.

Initial IP address is = 172.16.32.0

Wild card mask =0.0.15.255

by adding above values we can find the last IP address of ACL.

after addition

Final IP address is = 172.16.47.255

The options that are available with question, Only option between the range is 172.16.47.254. So we can say that This is the only ACE IP address in options.

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3 years ago

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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What are the primary functions of motor oil? a. Reduce friction and prevent wear b. Keep engine surfaces clean c. Remove heat to

ryzh [129]

The answer is E: all of the above.

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A mobile device user is having problems launching apps and texting. The user touches an app to launch it, but the app icon next

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Answer:

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