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solmaris [256]
1 year ago
11

Andrew must spend less than 53$ on meals during the weekend. he has already spent 21$ on meals costing 8$ average. how many addi

tional meals can andrew buy this weekend
Mathematics
1 answer:
Shkiper50 [21]1 year ago
3 0

Andrew can buy maximum 3 meals this weekend.

From given question,

Andrew must spend less than 53$ on meals during the weekend.

He has already spent 21$ on meals costing 8$ average.

Let x, the number of meals

So, we get an inequality,

8x + 21 < 53

We need to find the number of meals he can buy this weekend.

From above inequality,

⇒ 8x + 21 < 53

⇒ 8x < 53 - 21

⇒ 8x < 32

⇒ x < 4

This means, from 1 to 3 meals.

Therefore, Andrew can buy maximum 3 meals this weekend.

Learn more about an inequality here:

brainly.com/question/19003099

#SPJ4

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Answer:

Option d.

Step-by-step explanation:

we know that

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The curve is called the probability density function (abbreviated as pdf).

We use the symbol f(x) to represent the curve

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3 years ago
Add or subtract the polynomials<br> (8x3+3x-2)+(9x2-7+3x-2x3)
crimeas [40]

Answer:

10x^3+9x^2-9

Step-by-step explanation:

8x^3+2x-2+9x^2-7+2x-2x^3 - Combine like terms

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3 years ago
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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:

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