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oksian1 [2.3K]
3 years ago
11

So I have to compute the following but i don't know what they mean any help?

Mathematics
2 answers:
Gelneren [198K]3 years ago
7 0

Answer:

Step-by-step explanation:

This is a Combination (as in permutation vs combination) question the symbol (n r) refers to "n choose r". This is sometimes written as nCr

i.e the question is asking you to find how many combinations each will yield when you chose r items from n item without repetition and order does not matter.

I will only do the first question for you and you can just follow the same steps to solve the rest of the questions.

Recall that

nCr=\frac{n!}{(r!)(n-r)!}

Consider question a) we are given (5 1) or ₅C₁

we can see that n = 5 and r = 1

If we substitute this into the formula:

₅C₁ = (5!) / [ (1!)(5- 1)!]

= (5!) / [ (5- 1)!]

= (5!) / (4!)

= (5·4·3·2·1) / (4·3·2·1)

= 5

hence ₅C₁ = 5

timama [110]3 years ago
4 0

Answer:

(a) 5

(b) 10

(c) 35

(d) 28

(e) 9

(f) 21

Step-by-step explanation:

(\frac{n}{k} )=\frac{n!}{k!(n-k)!}

(a)

(\frac{5}{1} )=\frac{5!}{1!(5-1)!} \\\\(\frac{5}{1} )=\frac{5*4*3*2*1}{1*4!} \\\\(\frac{5}{1} )=\frac{120}{1*4*3*2*1} \\\\(\frac{5}{1} )=\frac{120}{24} \\\\(\frac{5}{1} )=5

(b)

(\frac{5}{3} )=\frac{5!}{3!(5-3)!} \\\\(\frac{5}{3} )=\frac{5*4*3*2*1}{3*2*1*2!} \\\\(\frac{5}{3} )=\frac{120}{3*2*1*2*1}\\\\(\frac{5}{3} )=\frac{120}{12} \\\\(\frac{5}{3} )=10

(c)

(\frac{7}{4} )=\frac{7!}{4!(7-4)!} \\\\(\frac{7}{4} )=\frac{7*6*5*4*3*2*1}{4*3*2*1*3!} \\\\(\frac{7}{4} )=\frac{5040}{4*3*2*1*3*2*1} \\\\(\frac{7}{4} )=\frac{5040}{144} \\\\(\frac{7}{4} )=35

(d)

(\frac{8}{2}) =\frac{8!}{2!(8-2)!} \\\\(\frac{8}{2}) =\frac{8*7*6*5*4*3*2*1}{2*1*6!} \\\\(\frac{8}{2}) =\frac{40320}{2*1*6*5*4*3*2*1}\\\\(\frac{8}{2}) =\frac{40320}{1440} \\\\(\frac{8}{2}) =28

(e)

(\frac{9}{8} )=\frac{9!}{8!(9-8)!} \\\\(\frac{9}{8} )=\frac{9*8*7*6*5*4*3*2*1}{8*7*6*5*4*3*2*1*1!} \\\\(\frac{9}{8} )=\frac{362880}{40320*1!} \\\\(\frac{9}{8} )=\frac{362880}{40320*1} \\\\(\frac{9}{8} )=\frac{362880}{40320} \\\\(\frac{9}{8} )=9

(f)

(\frac{10}{4} )=\frac{10!}{4!(10-4)!} \\\\(\frac{10}{4} )=\frac{10*9*8*7*6*5*4*3*2*1}{4*3*2*1*6!} \\\\(\frac{10}{4} )=\frac{3628800}{24*6*5*4*3*2*1} \\\\(\frac{10}{4} )=\frac{362880}{17280} \\\\(\frac{10}{4} )=21

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arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

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Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

7 0
3 years ago
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If Sandra mixes <em>x</em> L of 65% solution with <em>y</em> L of 90% solution, then the resulting mixture has a total volume of

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Solve the first equation for <em>y</em> :

<em>y</em> = 500 - <em>x</em>

<em />

Substitute this into the second equation and solve for <em>x</em> :

0.65<em>x</em> + 0.90 (500 - <em>x</em>) = 375

0.65<em>x</em> + 450 - 0.90<em>x</em> = 375

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<em>x</em> = 300

Solve for <em>y</em> :

<em>y</em> = 500 - 300

<em>y</em> = 200

So, Sandra should mix 300 L of 65% solution with 200 L of 90% solution.

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