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3 years ago

Please help me with number 6

Mathematics

2 answers:

Vlad [161]3 years ago

4 0

Answer:

The answer is 250 dollars. to find this do 25 times 1000. the placement is 25 over 100 and x as in the amount were trying find over 1000.

Step-by-step explanation:

son4ous [18]3 years ago

3 0

The answer is C) $250

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Solve 2 (x + 4) - 10 = 14 step by step

Blababa [14]

Answer:

x=8

Step-by-step explanation:

2(x+4) - 10 =14

distribute the 2 2x +8 -10 +14

simplify 2x-2=14

add 2 2x=16

devide by 2 x=8

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3 years ago

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aalyn [17]

Answer:

tan53=x/14

Step-by-step explanation:

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3 years ago

Show step by step In a day there are 86,400 seconds. Write this number in scientific notation.

Nesterboy [21]

Answer:

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Step-by-step explanation:

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3 years ago

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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

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

ipn [44]

Answer:

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

Step-by-step explanation:

Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.

So, let x be the variable describing the marks of a student, and

$z=(x-\mu)/\sigma$

the standardized equivalent of that (mu - mean, sigma - standard deviation).

We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:

z-score for the 10% point: z_10=1.28

z-score for the 20% point: z_20=-0.84

That gives us two equations:

$z_{10}=1.28=(75-\mu)/\sigma\\z_{20}=-0.84=(40-\mu)\sigma$

and can be solved for mu and sigma (do the work on your end, I am showing my result):

$\mu=53.87\,\,,\,\, \sigma=16.51$

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

3 0

3 years ago