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1 year ago

Find the number of permutations in the word “swimming”.

1 answer:

3 0

I believe it is 8!/(2!*2!)
You have a total of 8 letters but you have 2 letters that repeat twice each.
(8*7*6*5*4*3*2*1)/(2*1*2*1)
8*7*6*5*3*2 = 10,080 ways

You might be interested in

The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to agi

Ivanshal [37]

Answer:

P(X \le 250 ) = 0.7564 [/tex] , $P(X < 250 ) = 0.7564$ ,

$P(X < 300 ) = 0.09922$

$P(100 < X < 250 ) =0.644$

$x = 192.1$

Step-by-step explanation:

From the question we are told that

The value for $\alpha = 2.6$

The value for $\beta = 220$

Generally the Weibull distribution function is mathematically represented as

$F( x , \alpha , \beta ) = \left \{ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x < 0} \atop { 1- e^{-(\frac{x}{\beta } )^{\alpha } }}\ \ \ \ \ \ x \ge 0} \right$

Generally the probability that a specimen's lifetime is at most 250 is mathematically represented as

$P(X \le 250 ) = F(250, 2.7 , 220 )$

$P(X \le 250 )=1 - e^{- (\frac{250}{220} )^{2.7}}$

$P(X \le 250 ) = 1 - 0.2436$

$P(X \le 250 ) = 0.7564$

Generally the probability that a specimen's lifetime is less than 250

$P(X < 250 ) = F(250, 2.7 , 220 )$

[texP(X < 250 ) =1 - e^{- (\frac{250}{220} )^{2.7}}[/tex]

$P(X < 250 ) = 1 - 0.2436$

$P(X < 250 ) = 0.7564$

Generally the probability that a specimen's lifetime is more than 300

$P(X > 300 ) = 1- p(X \le 300 )$

$P(X > 300 ) = 1- F(300, 2.7 , 220 )$

[texP(X < 300) =1- [1 - e^{- (\frac{300}{220} )^{2.7}}][/tex]

$P(X < 300 ) = 0.09922$

Generally the probability that a specimen's lifetime is between 100 and 250 is

$P(100 < X < 250 ) = P(X < 250) - P(X < 100)$

=> $P(100 < X < 250 ) =F(250 , 2.7 , 220 ) - F(100 , 2.7 , 220 )$

=> $P(100 < X < 250 ) =(1 - e^{-(\frac{250}{220})^{2.7}}) - (1 - e^{-(\frac{100}{220})^{2.7}})$

=> $P(100 < X < 250 ) = (1 - 0.244 ) - (1- 0.888)$

=> $P(100 < X < 250 ) =0.644$

Generally the value such that exactly 50% of all specimens

$P(X > x) = 1-P(X < x) = 0.50$

=> $P(X > x) = 1- (1 - e^{- (\frac{x}{220}) ^{2.7}}) = 0.50$

=> $P(X> x ) = e^(- \frac{x}{20})^{2.7} = 0.50$

=> $P(X> x ) = (- \frac{x}{20})^{2.7} = ln0.50$

=> $P(X> x ) = \frac{x}{20} =[ -ln0.50 ] ^{frac{1}{2.7}}$

=> $x = 220[ -ln0.50 ] ^{frac{1}{2.7}}$

=> $x = 192.1$

7 0

1 year ago

Suppose that, for every lot of 100 computer chips a company produces, an average of 1.4 are defective. Another company buys many

Allisa [31]

Answer:

0.1665

Step-by-step explanation:

Given that on an average 1.4 are defective per 100 computer chips

If the tested lot contains more than three defects, the buyer will reject all the lots sent in that batch.

Let X be the no of defective items in 100 chips

X is Poisson with mean =1.4

i.e. the probability that the buyer will accept the lots

= $P(X\geq 3)\\\\=0.16651$

Answer is 0.1665

7 0

2 years ago

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago

The college has 700 students, and has issued student 388 parking permits. What is the percentage of students with parking permit

andreyandreev [35.5K]

55.4% ( to 1 decimal place )

to find percentage, make a fraction with parking permits issued on numerator and total students on denominator then multiply by 100%

percent of students with parking permits = $\frac{388}{700}$ × 100% = 55.4%

4 0

1 year ago

60% of the students in Mr. Vick's class have computers at home. If 21 students in his class have computers at home, how many stu

aleksley [76]

Answer:

There are 35 students in his class.

Step-by-step explanation:

1. Create a proportion: 60/100=21/x

2. Cross multiply, or not (I prefer to cross multiply). 60x=2100

3. Solve the equation: x=35.

Answer: There are 35 students in his class, 14 people going in school, 21 student online.

3 0

1 year ago