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3 years ago

A florist is making bouquets. Draw lines to match each

Mathematics

1 answer:

WITCHER [35]3 years ago

4 0

Answer:

1) 630 roses with 10 roses in each bouquet = 63 bouquets

2) 640 roses with 12 roses in each bouquet = 53 bouquets

3) 975 roses with 18 roses in each bouquet = 54 bouquets

4) 1,250 roses with 24 roses in each bouquet = 52 bouquets

Step-by-step explanation:

We need to match each description with the number of bouquets that can be made.

Solving this, we need to make bouquet of roses and each bouquet contains the given number of roses. We need to divide the total roses available by number of roses in each bouquet to find number of bouquets.

The formula used will be: $Number\:of\:bouquets=\frac{Total\;roses}{Number\:of\:roses\:in\:each\:bouquet}$

1) 630 roses with 10 roses in each bouquet

Total roses = 630

Number of roses in each bouquet = 10

So, number of bouquets will be:

$Number\:of\:bouquets=\frac{Total\;roses}{Number\:of\:roses\:in\:each\:bouquet}\\Number\:of\:bouquets=\frac{630}{10}\\Number\:of\:bouquets=63$

So, 63 bouquets can be made.

2) 640roses with 12 roses in each bouquet

Total roses = 640

Number of roses in each bouquet = 12

So, number of bouquets will be:

$Number\:of\:bouquets=\frac{Total\;roses}{Number\:of\:roses\:in\:each\:bouquet}\\Number\:of\:bouquets=\frac{640}{12}\\Number\:of\:bouquets=53.3 \approx53$

So, 53 bouquets can be made.

3) 975 roses with 18 roses in each bouquet

Total roses = 975

Number of roses in each bouquet = 18

So, number of bouquets will be:

$Number\:of\:bouquets=\frac{Total\;roses}{Number\:of\:roses\:in\:each\:bouquet}\\Number\:of\:bouquets=\frac{975}{18}\\Number\:of\:bouquets=54.1\approx54$

So, 54 bouquets can be made.

4) 1,250 roses with 24 roses in each bouquet

Total roses = 1250

Number of roses in each bouquet = 24

So, number of bouquets will be:

$Number\:of\:bouquets=\frac{Total\;roses}{Number\:of\:roses\:in\:each\:bouquet}\\Number\:of\:bouquets=\frac{1250}{24}\\Number\:of\:bouquets=52$

So, 52 bouquets can be made.

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Step-by-step explanation:

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3 years ago

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use 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.

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

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 $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

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

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

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

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

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

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

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