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Alla [95]
2 years ago
5

How much would you make for one month (4 weeks) if you worked a 40 hour week?

Mathematics
2 answers:
Anon25 [30]2 years ago
7 0

Answer:

Unclear

Step-by-step explanation:

I believe we need more information to solve the problem!

We currently have:

a  = b

b = total amount for 40 hours

As you can see, we cannot solve this. For a we would need something like how much you make per week, etc!

Kisachek [45]2 years ago
7 0
X=4hrs. X=4weeks. 2(4x4)=?
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A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January
Blababa [14]

Answer:

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

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

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

Step-by-step explanation:

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.

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

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

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

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

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

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

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

5 0
3 years ago
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