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grigory [225]
2 years ago
8

PLZ HELP HURRY PLZ!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Mademuasel [1]2 years ago
4 0

Answer:

D

Step-by-step explanation:

You subtract the "-6" from the y, and you get +6 (which eliminates A and B). You subtract the "-2" from the x, and you get +2 (which eliminates C)

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An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6
SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6  times.

The above situation can be represented through binomial distribution;

P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......

where, n = number trials (samples) taken = 6 trials

            r = number of success = at least 4

           p = probability of success which in our question is probability of

                 rolling a “six", i.e; p = 0.20

<u><em>Let X = Number of sixes on a die</em></u>

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X \geq 4)

P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)

=  \binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}

=  15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}

=  0.0154 + 0.00154 + 0.000064

=  0.01696

<em />

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0
3 years ago
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Step-by-step explanation:

solve it

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