How can this be solved?

Mathematics

2 answers:

Molodets [167]3 years ago

8 0

You might have to multiply or divide I just started learn g that stuff and I do t understand it still

svetoff [14.1K]3 years ago

8 0

You might have to multiply i don't think this a problem you use divison so multiply and then if im correct you then add hope this helps i have not had to do this in a while so hope it helps to find the answer

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TIME SENSITIVE! PLEASE HELP!

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Answer:

Answer choice A

Step-by-step explanation:

For the first letter, there are 26 options, as there are for the second and third letter. For each of the three numbers, there are 10 digits to choose from. Therefore, 26*26*26*10*10*10=17,576,000, or answer choice A. Hope this helps!

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3 years ago

Solve: x^2-x-6=0<br> Show your wrok

ivolga24 [154]

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2 years ago

A variation on a familiar theme: A packet is received correctly by an end user with probability p. To increase the likelihood th

Leya [2.2K]

Answer:

a) X can take the values {0,1,2,3,...,n}

The probability mass function of x is defined for k copies received correctly as:

$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$

b) N should be N=2 to ensure that with probability 0.99 at least one copy of the message is received.

Step-by-step explanation:

The variable X:number of copies received correctly by the end user can be modeled as a binomial variable, as it is a sum of n Bernoulli variables with probability p.

X can take the values {0,1,2,3,...,n}

The probability mass function of x is defined for k copies received correctly as:

$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$

being p: the probability that a packet is received correctly by an end user, and q=1-p: the probability that a packet is not received correctly by an end user.

b) We have to calculate n so as to have a probability P(x≥1)=0.99, when p=0.9.

$P(x\geq1)=1-P(x=0)=0.99\\\\\\P(x=0)=\dbinom{n}{0}p^0(1-p)^n=1*1*(1-p)^n=(1-0.9)^n=0.1^n\\\\\\P(x\geq1)=1-0.1^n=0.99\\\\0.1^n=1-0.99=0.01\\\\n\cdot ln(0.1)=ln(0.01)\\\\n=ln(0.01)/ln(0.1)=2$