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sveticcg [70]
3 years ago
9

Given the value x=false ,y=5 and z=1 what will be the value of F=(4%2)+2*Y +6/2 +(z&&x)?

Computers and Technology
1 answer:
Bezzdna [24]3 years ago
5 0
<h2>Answer:</h2>

F = 13

<h2>Explanation:</h2>

Given:

x = false

y = 5

z = 1

F = (4%2)+2*Y +6/2 +(z&&x)

We solve this arithmetic using the order of precedence:

<em>i. Solve the brackets first</em>

=> (4 % 2)

This means 4 modulus 2. This is the result of the remainder when 4 is divided by 2. Since there is no remainder when 4 is divided by 2, then

4 % 2 = 0

=> (z && x)

This means (1 && false). This is the result of using the AND operator. Remember that && means AND operator. This will return false (or 0) if one or both operands are false. It will return true (or 1) if both operands are true.

In this case since the right operand is a false, the result will be 0. i.e

(z && x) = (1 && false) = 0

<em>ii. Solve either the multiplication or division next whichever one comes first.</em>

=> 2 * y

This means the product of 2 and y ( = 5). This will give;

2 * y = 2 * 5 = 10

=> 6 / 2

This means the quotient of 6 and 2. This will give;

6 / 2 = 3

<em>iii. Now solve the addition by first substituting the values calculated earlier back into F.</em>

F = (4%2)+2*Y +6/2 +(z&&x)

F = 0 + 10 + 3 + 0

F = 13

Therefore, the value of F is 13

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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
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Answer:

a. 1365 ways

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c. Probability = 0.5904

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Given

PCs = 15

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^nC_r = \frac{n!}{(n-r)!r!}

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

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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

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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

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