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3 years ago

Using a table of values, determine the solution to the equation below to the nearest fourth of a unit. 2^x=1-3^x

Mathematics

1 answer:

Greeley [361]3 years ago

6 0

Answer:

Option (1)

Step-by-step explanation:

Given equation is,

$2^x=1-3^x$

To determine the solution of the equation we will substitute the values of 'x' given in the options,

Option (1)

For x = -0.75

$2^{-0.75}=1-3^{-0.75}$

0.59 = 1 - 0.44

0.59 = 0.56

Since, values on both the sides are approximately same.

Therefore, x = -0.75 will be the answer.

Option (2)

For x = -1.25

$2^{-1.25}=1-3^{-1.25}$

0.42 = 1 - 0.25

0.42 = 0.75

Which is not true.

Therefore, x = -1.25 is not the answer.

Option (3)

For x = 0.75

$2^{0.75}=1-3^{0.75}$

1.68 = 1 - 2.28

1.68 = -1.28

Which is not true.

Therefore, x = 0.75 is not the answer.

Option (4)

For x = 1.25

$2^{1.25}=1-3^{1.25}$

2.38 = 1 - 3.95

2.38 = -2.95

It's not true.

Therefore, x = 1.25 is not the answer.

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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: