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3 years ago

9

Once there was a girl with five lollipops. i took them because i did not eat breakfast. so how many are left. If you delete my q

uestion then i will ban you from life heh. Plz answe this is urgent.

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

The girl has 0 left. You on the other hand has around 4.99 to 5.01 lollipops

Step-by-step explanation:

thankfully they were poisoned lollipops that the girl was secretly hired to have you take them for you to eat.

. 。 • 　ﾟ。　. 　. . 　。。　. . 。 ඞ 。 . •The girl was an imposter. . . 。　. 　。ﾟ　. 　. , . . .

im sorry i had to put that

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Answer:

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Step-by-step explanation:

$12 = f - 13-2$

Collect like terms and simplify

$12+13+2 = f\\27 =f$

Switch sides

$f = 27$

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3 years ago

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Answer: x=-10, y=13

Step-by-step explanation:

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3 years ago

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3 years ago

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Answer:

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3 years ago

If 20% of the people in a community use the emergency room at a hospital in one year, find

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Answer:

a) 87.91% probability that at most three used the emergency room

b) 20.13% probability that exactly three used the emergency room.

c) 3.28% probability that at least five used the emergency room

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use the emergency room, or they do not. The probability of a person using the emergency room is independent of any other person. So we use the binomial probability distribution 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.

Sample of 10 people:

This means that $n = 10$

20% of the people in a community use the emergency room at a hospital in one year

This means that $p = 0.2$

a) At most three used the emergency room

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

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$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$

87.91% probability that at most three used the emergency room

b) Exactly three used the emergency room

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

20.13% probability that exactly three used the emergency room.

c) At least five used the emergency room

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

In which

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

From 0 to 3, we already have in a).

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

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$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9672 = 0.0328$

3.28% probability that at least five used the emergency room

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3 years ago