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3 years ago

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Lucas started off with $10.00. he gave away 20% of his money to his sister, and 30% of the remainder of his money to his brother

. How much money did Lucas have after sharing with his sister and his brother

2 answers:

Korolek [52]3 years ago

6 0

Answer:

5.60

Step-by-step explanation:

20% of 10 = 2 he gave 2 dollars to his sis

2 - 10 = 8 now lucas has 8 dollars

30% of 8 = 2.4 he gave 2.40 dollars to his bro

2.4 - 8 = 5.6

now he has 5 dollars and 60 cents

<u><em>give me brainlest pls pls pls :></em></u>

faltersainse [42]3 years ago

6 0

Answer:

5.60

Step-by-step explanation:

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a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

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d) The expected number is 2.3 and the standard deviation is 1.33.

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We use the binomial probability distribution to solve this question.

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

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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.

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And p is the probability of X happening.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

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

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

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c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

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$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

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