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borishaifa [10]
2 years ago
9

Considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we pr

esent their model. For our purposes, it can be described as a multinomial model with the four categories C1, C2, C3 , and C4 . For a sample of size n, let X = (X1, X2, X3 , X4 )′ denote the observed frequencies of the four categories. Hence, n = +4 i=1 Xi. The probability model is
Mathematics
1 answer:
Brrunno [24]2 years ago
4 0

Answer:

im not sure

Step-by-step explanation:

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Temka [501]
3sqrt(2)

You can either use Pythagorean theorem or special triangle proportions to find the side that is parallel to z. Since in a 45-45-90 degree triangle, the hypotenuse is the leg times the square root of 2, and the hypotenuse is 3, we know that the leg is 3/sqrt(2) which is equal to 3 sqrt(2) / 2. Since z is twice this length, we know that z is equal to 2 x 3sqrt(2)/2 = 3sqrt(2)
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2 years ago
HELP PLZ AND I WILL GIVE U BRAINLIEST
arsen [322]

Answer:

Step-by-step explanation:

1. A

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2 years ago
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azamat

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Step-by-step explanation:

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2 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
How many solutions does the following equation have? <br><br>-6(x+7) = -4x -2​
Fed [463]
Solve for x first
-6x - 42 = -4x -2
-6x = -4x + 40
- 2x = 40
X = -20
The equation has one solution
3 0
2 years ago
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