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3 years ago

Witch of the following expression is equivalent to the logarithmic expression below. log(3)5/x^2

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

5 0

Answer: The correct option is

(B) $\log_35-2\log_3x.$

Step-by-step explanation: We are given to select the expression that is equivalent to the following logarithmic expression :

$E=\log_3\dfrac{5}{x^2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following properties of logarithms :

$(i)~\log_a\dfrac{b}{c}=\log_ab-\log_ac,\\\\(ii)~\log_ab^c=c\log_ab.$

From (i), we get

$E\\\\=\log_3\dfrac{5}{x^2}\\\\\\=\log_35-\log_3x^2~~~~~~~~~~~~~~~~~~~~[\textup{Using property (i)}]\\\\\\=\log_35-2\log_3x.~~~~~~~~~~~~~~~~~~~~[\textup{Using property (ii)}]$

Thus, the required equivalent expression is $\log_35-2\log_3x.$

Option (B) is CORRECT.

BabaBlast [244]3 years ago

4 0

Assuing you mean
$log_{3}(\frac{5}{x^2})$
remember that $log(\frac{a}{b})=log(a)-log(b)$
also, $log(x^m)=(m)log(x)$

so
$log_{3}(\frac{5}{x^2})=$
$log_{3}(5)-log_{3}(x^2)=$
$log_{3}(5)-2log{3}(x)$

the answer is B

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3 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c: