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Vadim26 [7]
3 years ago
12

Can someone please help me with this question?

Computers and Technology
1 answer:
madreJ [45]3 years ago
5 0
I would use a biometric machine that scans a fingerprint and a eye retina then calls up my bank information from the scan. Who wants to carry a card around anyway?
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Write a code that takes numbers from the user as a list. (User can enter as many numbers as he wants). Then, find mean and stand
8090 [49]

nums = list(map(int, input().split()))

mean = sum(nums)/len(nums)

print("The mean is",mean)

values = 0

for x in nums:

   values += abs(x - mean)**2

print("The standard deviation is",(values / (len(nums)-1))**0.5)

I wrote my code in python 3.8. I hope this helps.

7 0
3 years ago
Firefox, Chrome, Opera, and Safari are examples of
MrRissso [65]
D. Browsers

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3 years ago
4.2 lesson practice help plzs
sertanlavr [38]

The output is 21 because c is incremented by 3 until it surpasses 10 and the value of c is added to sum each time the loop runs.

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8 0
2 years ago
Which principle suggests that specific single responsibility interfaces are better than one general purpose interface?
ElenaW [278]

Answer:Interface segregation principle

Explanation: Interface-segregation principle (ISP) is amongst the major principles of the object-oriented design which describes that none of the users/clients can be forced for indulging and depending on the unknown methods or methods that they don't have knowledge about.

It functions by making the interfaces visible to the user that specifically fascinates them and keeping other smaller interfaces.Interfaces are made by splitting process and making the small interfaces from them.

8 0
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

<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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