How many 3-person committees can be formed in a club with 8 members??

Mathematics

1 answer:

earnstyle [38]3 years ago

7 0

=8C3
=8!/(3!*5!)
=56 ways

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the mean average distance that jack runs in his last 10 runs is 5.7 miles. work out the distance that he would need to make this

ziro4ka [17]

Answer:

9 miles

Step-by-step explanation:

mean is calculated as

mean = $\frac{sum}{count}$ , then

$\frac{sum}{10}$ = 5.7 ( multiply both sides by 10 )

sum = 57

let x be the distance he needs to cover so mean is 6 , that is

$\frac{sum+x}{11}$ = 6 ( multiply both sides by 11 )

sum + x = 66 , that is

57 + x = 66 ( subtract 57 from both sides )

x = 9

That is he would have to run 9 miles

3 0

2 years ago

[6+4=10 points] Problem 2. Suppose that there are k people in a party with the following PMF: • k = 5 with probability 1 4 • k =

kirza4 [7]

Answer:

1). 0.903547

2). 0.275617

Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = $1-\frac{_{n}^{12}\textrm{P}}{12^2}$

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= $\frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})$