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Mrrafil [7]
2 years ago
15

"A consultant compiled the following data set that shows" the number of visits made to the National Museum of American History f

rom 2001 to 2015. The consultant noticed that the number of visits in 2007 and 2008 seemed unusually low compared to the rest of the data set. What should the consultant do about the data points from 2007 and 2008?Year Visits2015 4,100,0002014 4,000,0002013 4,900,0002012 4,800,0002011 4,600,0002010 4,200,0002009 4,400,0002008 480,0002007 02006 2,400,0002005 3,000,0002004 2,900,0002003 2,600,0002002 4,200,0002001 5,200,000
Mathematics
1 answer:
aalyn [17]2 years ago
3 0

Answer:

Step-by-step explanation:

The best option is for the consultant to remove these data points because they are outliers.  Unusual data points which are located far from rest of the data points are known as outliers.

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Assume there are 365 days in a year.
MissTica

Answer:

1) The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.

Step-by-step explanation:

Given : Assume there are 365 days in a year.

To find : 1) What is the probability that ten students in a class have different birthdays?

2) What is the probability that among ten students in a class, at least two of them share a birthday?

Solution :

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

Total outcome = 365

1) Probability that ten students in a class have different birthdays is

The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883

The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday

P(2 born on same day) = 1- P( 2 not born on same day)

\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]

\text{P(2 born on same day) }=1-[\frac{364}{365}]

\text{P(2 born on same day) }=0.002

The probability that among ten students in a class, at least two of them share a birthday is 0.002.

6 0
2 years ago
Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in
Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

f_n=5f_{n-4}+3f_{n-5}

for n=5,6,7,..

Initial condition

f_0=0,f_1=1,f_2=1,f_3=2,f_4=3

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

f_n=f_{n-1}+f_{n-2},f_0=0,f_1=1

Apply this

f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}

f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}

f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}

Substitute n=2

f_2=f_1+f_0=1+0=1

f_3=f_2+f_1=1+1=2

f_4=f_3+f_2=2+1=3

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that f_{5n} is divisible by 5 for n=1,2,3,..

Now replace n by 5n

f_{5n}=5f_{5n-4}+3f_{5n-5}

Apply induction

Substitute n=1

f_5=5f_1+3f_0=5+0=5

It is true for n=1

Suppose it is true for n=k

f_{5k}=5f_{5k-4}+3f_{5k-5} is divisible 5

Let f_{5k}=5q

Now, we shall prove that for n=k+1 is true

f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)

f_{5k+5}=5(f_{5k+1}+3q)

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the f_{5n} is divisible by 5 for n=1,2,3,..

8 0
1 year ago
The product of a binomial and a trinomial is x 3 + 3 x 2 − x + 2 x 2 + 6 x − 2. Which expression is equivalent to this product a
Mazyrski [523]
X3+5x2+5×-2 is the correct answer.
4 0
1 year ago
Read 2 more answers
The tax rate as a percent, r, charged on an item can be determined using the formula StartFraction c Over p EndFraction minus 1
Nataliya [291]

Answer:

43.20

Step-by-step explanation:

Had to take one for the team

4 0
1 year ago
Read 2 more answers
A sociologist studying the difference in ages between husbands and wives obtained a random sample of 55 married couples. The mea
zalisa [80]

Answer:

2.1/√55

Step-by-step explanation:

simga divided by sample size

8 0
2 years ago
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