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REY [17]
4 years ago
9

A boy wins a carnial games 50% of the time. He uses a spinnser with equal-sized sections labeled 0 - 9 to simulate trials of the

game. In the simulation, the numbers 0-4 represent winnning. Numbers 5-9, represent losing. Based on the simulations below that represent the boy playing the game 3 times, what is the probability he will win all 3 times? HINT: Your answer must be written in fraction: Favorable Outcome/Total Possible Outcome.
412 750 236 808 904 237 424 648 121 208
Mathematics
2 answers:
Alenkasestr [34]4 years ago
8 0

Answer:

412,424,121

Step-by-step explanation:

3/10 equals 309.

Nina [5.8K]4 years ago
5 0

Answer:

is it one of the numbers down below it? Cuase if it is then probablay 648

Step-by-step explanation:

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stiv31 [10]

Answer:

A. 0.2, .25, 0.2

B. Daniella and Angel have tied scores, but Dwayne had the highest of all three by .05 (5%)

Step-by-step explanation:

6 0
3 years ago
Determine the value of a in the figure shown
galina1969 [7]

Answer:

d

Step-by-step explanation:

180-149= 31.

31+ 122= 153

180-153= 27

a= 27

3 0
3 years ago
I need help but quick tho
Ivan

Answer:

D.

Step-by-step explanation:

x  =  ±√36

x  =  ±6

hopes this helps :)

7 0
3 years ago
Read 2 more answers
What are the solutions to the quadratic equation (5y + 6)2 = 24?
nata0808 [166]

Answer:

see explanation

Step-by-step explanation:

Given

(5y + 6)² = 24 ( take the square root of both sides )

5y + 6 = ± \sqrt{24} = ± 2\sqrt{6} ( subtract 6 from both sides )

5y = - 6 ± 2\sqrt{6} ( divide both sides by 5 )

y = \frac{-6+2\sqrt{6} }{5} and y = \frac{-6-2\sqrt{6} }{5}

3 0
3 years ago
Let X1, · · · , Xn be independent identically distributed random variables with probability density function f(x) = 1 2σ exp − |
Igoryamba

Answer:

σˆ =

sPn

i=1 X2

i

2n

Step-by-step explanation:

To obtain, the maximum likelihood ratio, we use the following method.

l(σ) = Xn

i=1 "

− log 2 − log σ −

|Xi

|

σ

#

Let the derivative with respect to θ be zero:

l

0

(σ) = Xn

i=1 "

−

1

σ

+

|Xi

|

σ

2

#

= −

n

σ

+

Pn

i=1 |Xi

|

σ

2

= 0

and this gives us the MLE for σ as

σˆ =

Pn

i=1 |Xi

|

n

Again this is different from the method of moment estimation which is

σˆ =

sPn

i=1 X2

i

2n

As our answer

8 0
3 years ago
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