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3 years ago

Why does world of tanks keep downloading when I'm not even at the computer

Computers and Technology

1 answer:

Lynna [10]3 years ago

8 0

Its the option, auto update on
You can turn it off in the launcher (before you get into the login screen)

You might be interested in

The operations planning practice of inputting sales forecasts into computer software that accurately predicts the amount and tim

netineya [11]

Answer:

"Materials requirement planning" is the correct answer for the above question.

Explanation:

The "Materials requirement planning" is a software system that is used to hold the record of the raw materials.
It is used to tell about the material which is present in the stocks. It also used to schedule the delivery.
This software is used to enhance the productivity of the company.
The above question asked about the software which is needed to keep the record of the raw material. This software is known as "Materials requirement planning".

3 0

3 years ago

16.

dimaraw [331]

Answer:

r = 15 cm

Explanation:

formula

V = πr²×h/3

replace

4950 = 22/7×r²×21/3

4950 = 22/7×r²×7

4950 = 22×r²

r² = 4950/22

r² = 225

r = √225

r = √15²

r = 15 cm

3 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

Pls answer will give brainlest dont answer if you dont know Upload your 300-word essay containing the following: the definition

Rainbow [258]

Answer: oh hello human :) also PLEASE DONT KILL ME I’m answering this because the question was already answered already on another post

6 0

3 years ago

In python please!! Write the definition of a function named countPos that needs integer values from standard input until there a

jek_recluse [69]

Answer:

Explanation:

The following code is written in Python it doesn't use any loops, instead it uses a recursive function in order to continue asking the user for the inputs and count the number of positive values. If anything other than a number is passed it automatically ends the program.

def countPos(number=input("Enter number: "), counter=0):

try:

number = int(number)

if number > 0:

counter += 1

newNumber = input("Enter number: ")

return countPos(newNumber, counter)

else:

newNumber = input("Enter number: ")

return countPos(newNumber, counter)

except:

print(counter)

print("Program Finished")

countPos()

8 0

3 years ago