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3 years ago

I have no idea how to do Log problems like this. Can anyone help me out?

Mathematics

2 answers:

UkoKoshka [18]3 years ago

6 0

I'm using the following rules:

Rule 1: log(x*y) = log(x)+log(y)
Rule 2: log(x/y) = log(x)-log(y)
Rule 3: log(x^y) = y*log(x)
Rule 4: sqrt(x) = x^(1/2)

log[ (x^2*y)/(sqrt(z)) ] = log[ x^2*y ] - log[ sqrt(z) ] .... Use Rule 2
log[ (x^2*y)/(sqrt(z)) ] = log[ x^2 ] + log[ y ] - log[ sqrt(z) ] .... Use Rule 1
log[ (x^2*y)/(sqrt(z)) ] = log[ x^2 ] + log[ y ] - log[ z^(1/2) ] .... Use Rule 4
log[ (x^2*y)/(sqrt(z)) ] = 2*log[ x ] + log[ y ] - (1/2)*log[ z ] .... Use Rule 3
log[ (x^2*y)/(sqrt(z)) ] = 2*a + b - (1/2)*c .... Use Substitution

So
log[ (x^2*y)/(sqrt(z)) ]
will break down into
2*log[ x ] + log[ y ] - (1/2)*log[ z ]
which is equivalent to
2a + b - (1/2)c
when a = log(x), b = log(y), c = log(z)

-----------------------------------------------------

So that's why the answer is choice (4) 2a + b - (1/2)c

Virty [35]3 years ago

4 0

Hello there!

The correct answer is option 4

We need to use the rules of logarithms to expand.

I hope this helps!

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Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

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$P(X)=(nCx)(p)^x (1-p)^{n-x}$

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Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

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Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

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