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10 months ago

// This pseudocode is intended to describe

1 answer:

ddd [48]10 months ago

3 0

Applying given information of a computational language in pseudocode, a code may be written to represent computing the price of an item on sale for 10% off.

<h3>What is a pseudocode?</h3>

A pseudocode can be considered as a depiction of the steps represented in an algorithm, usually in plain (natural) language.

//This pseudocode is intended to describe

//computing the price of an item on sale for 10% off

START

input origPrice

discount = origPrice * 0.10

finalPrice = origPrice - discount

output finalPrice

STOP

//This pseudocode is intended to compute the number

//of miles per gallon you get with your car.

START

input milesTraveled

input gallonsOfGasUsed

milesPerGallon = milesTraveled / gallonsOfGasUsed

//milesPerGallon is computed using division

output milesPerGallon

//miles is misspelled, and the P in milesPerGallon should be uppercase

STOP

//Program should end with stop

//This pseudocode is intended to describe

//computing the per day cost of your rent

//in a 30-day month

START

input rent

costPerDay = rent / 30

// Comment indicates 31-day month

output costPerDay

// output should be costPerDay

STOP

Learn more about the pseudocode here :

brainly.com/question/13208346

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3 years ago

Discribe two ways you can zoom in and out from an image

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Answer:

Select the Zoom tool, and then do any of the following:

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<h2>psst! pls, brailiest!</h2>

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3 years ago

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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

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

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2 years ago

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andriy [413]

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