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lozanna [386]
3 years ago
7

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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320+168=$ 488

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x - 9 < 4
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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:

  • One person, thus q = 1.

The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

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

A similar problem is given at brainly.com/question/24756209

3 0
1 year ago
AS one of her chores Staci must vacum 1/2 of the house. If she has already vacumed 1/3 of her share tonight whatp part of the li
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Answer:

1/6

Step-by-step explanation:

sim[le thought


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