What would the range(3, 9) function generate?

This is going to get taken down but I dont care add me on discord cause ima bored -Red-#9847

Alex777 [14]

Answer:

I WILL

Explanation:

6 0

3 years ago

To ____ a public member function of a base class in the derived class, the corresponding function in the derived class must have

Norma-Jean [14]

Answer:

None of the mentioned options

Explanation:

If we define a method in a derived class with same declaration type as that of base class then it is said to be overriding a function which behaves differently which will depend on the object which is calling the method.
Option a,b and c are not any kind of methods we implement using coding.
Option d is done when number or data types of parameters are different than the declaration of the base type.

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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

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

8 0

3 years ago

Green Fields Landscaping Company sells evergreen trees which are priced by height. Customers have a choice of purchasing a tree

Tomtit [17]

Answer:

Answered below

Explanation:

//Program is written in Python programming //language.

number_of_trees = int(input ("Enter number of trees purchased: "))

height_of_trees = float(input("Enter height of trees: "))

delivery_status = input("Do you want trees delivered? enter yes or no ")

price_of_two_meters = 20

total_price = number_of_trees * price_of_two_meters

//Invoice

print (number_of_trees)

print(height_of_trees)

print (total_price)

print (delivery_status)

8 0

3 years ago

What is a distraction that you find when photographing animals at the zoo

yaroslaw [1]

D. All of the above.

Reason:

A. Fences can get in the way of a beautiful picture.

B: The animal's enclosure may be hard to get the perfect snap. Like a rock can be in the way of the animal. idk...lol

C. People want to see the animal as much as you do so they may block the way of a picture.

Example from Google..XD

7 0

3 years ago