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Nataliya [291]
3 years ago
13

Anyone ...Pls solve

Mathematics
1 answer:
expeople1 [14]3 years ago
6 0
Given:
- Bought n items of product A at 1/5 rs each
- bought n items of product B at 1/4 rs each
He mixed the two items and sold each at 2/9 rs each.
He lost 3 rs in the above batch of articles.

Solution:
revenue = 2n* (2/9) rs = 4n/9 rs
cost = n/5+n/4 rs = n(1/5+1/4)
Profit = revenue-cost = 4n/9-n(1/5+1/4) = -3 rs

Solve for n
4n/9-n(1/5+1/4) = -3 
factor out n
n(4/9-1/5-1/4)=-3
n(80-36-45)/180=-3
-n/180=-3
n=3*180=540

Total number of products bought
= n+n = 540+540 = 1080
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2 years ago
Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s
notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

No-show rate of 10%.

This means that p = 0.1

Four travelers who have made hotel reservations in this study.

This means that n = 4

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561

P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

X = 0 has the highest probability, which means that 0 is the most likely value for X.

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Answer:

Quadratic

Step-by-step explanation:

Linear has a constant growth and exponential cannot be negative.

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Answer:

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