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AfilCa [17]
4 years ago
12

I am stuck between two

Mathematics
2 answers:
Zarrin [17]4 years ago
5 0

Answer:

14

Step-by-step explanation:

-12/6 is -2

-2x-3 is 6

6+8 Is 14

Marina86 [1]4 years ago
3 0

Answer:

It's 14.

Step-by-step explanation:

Google. ;)

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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}


In which

x is the number of sucesses


e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial 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.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164


0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671

0.671 = 67.1% probability that neither contains a missing pulse

8 0
3 years ago
Share £15 in 2:3 <br> Please help me put
son4ous [18]
2 ratio 3,so, 2+3=5
15/5=3 add and then division
5 0
3 years ago
During the first week of a summer camp 2 out of 3 campers were boys. During the second week 3 out of 5 were boys. There were a 1
rewona [7]

Step-by-step explanation:

2/3=10/15 (divide 15 by 3 to get 5, then multiply both numerator and denominator by 5 to set equivalent)

3/5=9/15

10/15 is greater than 9/15, so during the 1st week there were more boy campers

6 0
3 years ago
What's pi to as many decimal places?
Reil [10]
2 decimal places for regular . 3.14
4 0
3 years ago
When converted to a fraction and reduced to simplest terms, which number will have
maks197457 [2]

Answer: 0.720

Step-by-step explanation:

0.720 = 18/25

0.51 = 51/100

0.144 = 18/125

0.023 = 23/1000

6 0
2 years ago
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