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max2010maxim [7]

2 years ago

8

Fourteen less children’s meals were served than adult meals at a barbecue. Children plate were 1.50 each and adult plates were 2

.00 each . If the total amount of money collected was 441 , how much of each type of plate was served

1 answer:

Annette [7]2 years ago

7 0

Answer:

134 children and 120 adult plates were served

Step-by-step explanation:

let the children mean be x

Let the adult meal be y

If fourteen less children’s meals were served than adult meals at a barbecue, then;

y = x - 14 .... 1

IF Children plate were 1.50 each and adult plates were 2.00 each with a total of 441 in amount then;

1.5x + 2y = 441 .... 2

Substitute 1 into 2;

1.5x + 2(x-14) = 441

1.5x+2x-28 = 441

3.5x = 441+28

3.5x = 469

x = 469/3.5

x = 134

Recall that y = x - 14

y = 134-14

y = 120

Hence 134 children and 120 adult plates were served

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Answer:

a) $P(X = 1) = 0.38742$

b) $P(X = 3) = 0.05740$

c) $P(X = 9) = 0.00000$

d) $P(X \geq 5) = 0.00163$

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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.

In this problem

There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

So

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

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