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eduard
3 years ago
14

Assign decoded_tweet with user_tweet, replacing any occurrence of 'TTYL' with 'talk to you later'. Sample output with input: 'Go

tta go. I will TTYL.' Gotta go. I will talk to you later.

Computers and Technology
2 answers:
nekit [7.7K]3 years ago
8 0

Answer:

I am going to use the Python programming language to answer this. The source code is given below:

print("Enter your tweet here")

user_tweet = input()

decoded_tweet = user_tweet.replace('TTYL', 'talk to you later')

print("This is the decoded tweet: ")

print(decoded_tweet)

Explanation:

In the program the replace() module was used to replace 'TTYL' with 'talk to you later.'

Attached is the screenshot of the output.

stich3 [128]3 years ago
4 0

Answer:

I am going to use the Python programming language to answer this. The source code is given below:

print("Enter your tweet here")

user_tweet = input()

decoded_tweet = user_tweet.replace('TTYL', 'talk to you later')

print("This is the decoded tweet: ")

print(decoded_tweet)

Explanation:

In the program the replace() module was used to replace 'TTYL' with 'talk to you later.'

Attached is the screenshot of the output.

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

<em>Hence, there are 1365 ways </em>

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)

<em>From the given parameters; 3 out of 15 is detective;</em>

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;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

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

8 0
3 years ago
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