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3 years ago

How do I create a flowchart to output numbers from n...1? We were doing loops during the classes.

Computers and Technology

1 answer:

Andrew [12]3 years ago

8 0

I use GOxxOGLE FORMS or WxxORDS and their good to make flow charts

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Please answer fast screenshot included - thanks in advance

aleksley [76]

Answer:

Webpage Layouts where finally designed using html code

Explanation:

4 0

3 years ago

Chat messages are most likely to be found where on a computer? firewall nic internet history ram

miv72 [106K]

The answer is RAM. Chat messages are most likely to be found in the RAM of the computer. It is Random Access Memory. RAM is considered volatile memory, which means that the stored information is lost when there is no more power.

3 0

3 years ago

Write a method that takes two circles, and returns the sum of the areas of the circles.

11111nata11111 [884]

Answer:

public static double areaSum(Circle c1, Circle c2){

double c1Radius = c1.getRadius();

double c2Radius = c2.getRadius();

return Math.PI * (Math.pow(c1Radius, 2) + Math.pow(c2Radius, 2));

public static void main(String[] args){

Circle c1 = new Circle(6.0);

Circle c2 = new Circle(8.0);

areaSum(c1,c2);

}

Explanation:

7 0

2 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

To pinpoint an earthquake's location, scientists need information from how many seismometers?

Mazyrski [523]

C. 3
Due to the different speeds of P and S waves, a single seismometers can determine the distance to an earthquake. So, for a single station, the localization is any point on a circle around the station. With 2 stations, you'll have two circles that intersect at two points. The 3rd station is needed in order to determine which of the 2 points is the actual earthquake.

7 0

3 years ago