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dybincka [34]
3 years ago
13

in any given year a factory has a 10% probability of having a serious accident. about every how many years might the factory exp

ect to have a accident​
Mathematics
1 answer:
Likurg_2 [28]3 years ago
5 0

Years for one serious accident is 10 year

<u>Given:</u>

Probability of having a serious accident = 10% every year

P(serious accident) = 0.1

Number of years = n

<u>Computation:</u>

Years for one serious accident = 1 / Probability of having a serious accident

Years for one serious accident = 1 / P(serious accident)

Years for one serious accident = 1 / 0.1

Years for one serious accident = 10 year

Learn more:

brainly.com/question/23044118?referrer=searchResults

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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)!}&#10;

In which

x is the number of sucesses

&#10;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&#10;

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
Rachel reads 1/5 of the pages of a book plus 12 pages on the first day.
kkurt [141]

Answer:

Let's say that the whole book have 5a pages.

First day: a + 12

Second day: (4a)1/4+15= a + 15

Third day: (1/3)[ 4a - (a+15)] +18= a + 13

Forth day: 70

So we get this:

(a + 12) + (a + 15) + (a + 13) + 70 = 5a

a + 12 + a + 15 + a + 13 + 70 = 5a

3a + 110 = 5a

3a - 5a = -110

-2a = -110

a = 55

5a = 5(55) = 275

So the total pages are in the book is:

275 pages

6 0
3 years ago
Read 2 more answers
Find the slope of the line passing through the pair points (7,6) and (-7,-6)
True [87]

Answer:

=6/7

Step-by-step explanation:

6 0
2 years ago
In an all boys school, the heights of the student body are normally distributed with a mean of 69 inches and a standard deviatio
slega [8]

Answer:

Two intervals are 63 and 75

Step-by-step explanation:

Given:

Mean of students (μ) = 69 inches

Standard deviation (σ) = 3 inches μ

Find:

Interval of heights (95%) using empirical rule

Computation:

Interval of heights (95%) using empirical rule:

⇒ μ – 2σ = 69 - 2(3)

⇒ μ – 2σ = 69 - 6

⇒ μ – 2σ = 63

⇒ μ + 2σ = 69 + 2 (3)

⇒ μ + 2σ = 69 + 6

⇒ μ + 2σ =  75

Two intervals are 63 and 75

7 0
3 years ago
A , B , C , Or D?!? :))))
sesenic [268]
The smaller triangle has a side of 16, while the corresponding side for the larger triangle is 64. 

The ratio of the two sides (small to large) is 16:64 which reduces to 1:4 after we divide both parts by 16. 

The ratio 1:4 basically means "multiply the smaller length by 4 to get the larger length". In other words, the larger length is 4 times longer. 

----------------------------------

From here, we square both parts of the ratio above
we go from 1:4 to 1:16 after we square both parts

What does the ratio 1:16 mean? It means that the larger area is 16 times that of the smaller area

If 
x = smaller area
y = larger area
then,
y = 16*x

----------------------------------

We're told that the larger area is 1360 square cm. So we know that y = 1360. Use this to find x

y = 16*x
1360 = 16*x
1360/16 = 16*x/16
85 = x
x = 85

----------------------------------

Answer: The smaller area is 85 square cm
Answer is choice D
7 0
3 years ago
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