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3 years ago

Least common multiple of 27.

Mathematics

1 answer:

DaniilM [7]3 years ago

8 0

Answer:

the least common multiple of 27 and 42 is 378

Step-by-step explanation:

(27*42)/3

1134/3

378

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Teacher was comparing two sets of quis scores shown below.

Bess [88]

Don’t listen to those people that put those things down to say open the file it’s a scam

8 0

3 years ago

) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.

yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the 0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= $\dfrac{1}{2}$

The length of bits : n = 10

Let X = Number of getting ones.

Then , $X \sim Bin(n=10,\ p=\dfrac{1}{2})$

Binomial distribution formula : $P(X=x)=^nC_x p^x q^{n-x}$ , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , $p=q=\dfrac{1}{2}$

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

$P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}$

$=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}$

$=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})$

$=(\dfrac{1}{2})^{10}(210+252)$

$=(0.0009765625)(462)$

$=0.451171875\approx0.4512$

Hence, the probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0

4 years ago

ANy help i'll give brainliest

Rus_ich [418]

Answer:

52o I think

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Urgent!!

suter [353]

Answer:

$70.62

Step-by-step explanation:

110 x 0.6=66

66x0.07 + 66=$70.62

3 0

3 years ago

(10 points) From a group of 9 men and 7 women a committee consisting of 4 men and 4 women is to be formed. How many different co

alexgriva [62]

Answer:

a) 2450

b) 1890

c) 1050

Step-by-step explanation:

Given:

- Available Group = 9 M & 7 F

- Committee of= 4 M and 4 W to be formed

Find:

- (a) 2 of the men refuse to serve together?

- (a) 2 of the women refuse to serve together?

- (a) 1 man and 1 woman refuse to serve together?

Solution:

- The question pertains to a selection process, We will use combinations for each case as follows.

part a)

- Since 2 men cant be selected together. Then we have to discount for one man while selecting Man for the committee. Hence, the combinations are:

8C4 x 7C4 = 2450

part b)

- Since 2 women cant be selected together. Then we have to discount for one woman while selecting them for the committee. Hence, the combinations are:

9C4 x 6C4 = 1890

part b)

- Since 1 women and a man cant be selected together. Then we have to discount for one woman and one man while selecting them for the committee. Hence, the combinations are:

8C4 x 6C4 = 1050

4 0

4 years ago