Only put coolant into your radiator when the engine is A. on B. hot C. warm D. cool

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

2 years ago

A(n) ________ is a chart based on PivotTable data.

Mrrafil [7]

Well since it’s a chart based on a PivotTable prettyyyy sure it’s gonna be a PibltChart

7 0

2 years ago

This question is related to linked-list representation of disjoint sets. Consider two sets Sx and Sy that have 3 and 5 elements,

sattari [20]

Answer:

A union (UNION(x,y)) of the sets Sx and Sy represented by x and y, respectively will perform ________4_________ update(s) of the attribute.

Explanation:

The UNION (x, y) disjoint-set data structure unites the dynamic sets that contain x and y, say Sx and Sy, into a new set. It is called the union of the two sets. Before the union operation, the two sets are disjoint. After the union operation, the representative of the resulting set is some member of Sx and Sy or either Sx or Sy. The sets Sx and Sy are then destroyed to remove them from the union collection S. So, four operations are required.

8 0

3 years ago

PLZZZZZ HELP!!!!!!!!!!

TiliK225 [7]

I'd say the best option would be B.

7 0

3 years ago

2. Which of the following is not one of the guidelines for using instant messaging?A.You can use in place of all face-to-face co

Alexxandr [17]

Answer:

A. You can use in place of all face-to-face communication.

Explanation:

Hope this helps

6 0

3 years ago