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2 years ago

What is log Subscript 15 Baseline 2 cubed rewritten using the power property?

Mathematics

2 answers:

nalin [4]2 years ago

7 0

Answer:

The equivalent expression

Step-by-step explanation:

Assuming your expression is $\log_{15}2^3$ , then we can use the power property of logarithm , which is $\log_ab^n=n\log_ab$ .

If we let a=15, b=2, and n=3

Then using the power property of logarithm will give us:

$\log_{15}2^3=3\log_{15}2$

Using the power property of logarithm, the equivalent expression is $3\log_{15}2$

garri49 [273]2 years ago

5 0

Answer:

Step-by-step explanation:

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The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the

navik [9.2K]

Answer:

a) 0.019

b) 0.563

c) x = 1.966 hours

Step-by-step explanation:

E(X) = 1

Exponential random variable's probability function is given as

P(X=x) = λ e^(-λ.x)

The cumulative distribution function is given as

P(X ≤ x) = 1 - e^(-λ.x)

a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.

But the number of planes that land every hour will be obtained using the Poisson distribution formula.

It is the best for discrete systems.

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = 1 aircraft per hour.

The probability that more than three aircraft arrive within an hour = P(X > 3)

P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p

b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals

Probability of one 1-hour interval not containing more than 3 arrivals = 1 - P(X > 3)

= 1 - 0.01899 = 0.98101

Probability that thirty 1-hour intervals will not contain more than 3 arrivals = (0.98101)³⁰ = 0.5626 = 0.563 to 3 d.p

c) Determine the length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14

We can now use the cumulative distribution function for exponential random variable for this

P(X ≤ x) = 1 - e^(-λ.x)

P(X > x) = 1 - P(X ≤ x)

P(X > x) = e^(-λ.x)

λ = 1, x = ?,

0.14 = e⁻ˣ

e⁻ˣ = 0.14

In e⁻ˣ = In 0.14 = -1.966

-x = -1.966

x = 1.966 hours

Hope this Helps!!!

4 0

2 years ago

Davis has 141 figures in his action figure collection. One-third of the figures are sports related. How many of the figures in D

Lelu [443]

We have been given that Davis has 141 figures in his action figure collection. One-third of the figures are sports related.

To find the number of figures that are not sports related we will find 2/3 of 141 as 2/3 of the figures will be not related to sports.

$\text{ Figures not related to sports}=141\times\frac{2}{3}$

$\text{ Figures not related to sports}=47\times2$

$\text{ Figures not related to sports}=94$

Therefore, 94 figures in Davis’ collection are not sports related.

7 0

2 years ago

Lou started his savings account with 50$ he now only has 31$. What percent of change did his savings account have?

vladimir1956 [14]

Answer:

Step-by-step explanation:

50 - 31 = 19

38%

4 0

2 years ago

A folk band sold CDs at a concert. Each single CD costs $12 and their double CDs cost $17. They sold 14 more single CDs than dou

attashe74 [19]

The band sold 33 single CDs and 19 double CDs.

33 x 12 = 396

19 x 17 = 323

396 + 323 = 719

7 0

2 years ago

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

Dovator [93]

Option a: $\frac{m^{5} }{162n}$ is the equivalent expression.

Explanation:

The expression is $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ where $m\neq 0, n\neq 0$

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

$\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }$

Cancelling the like terms, we have,

$\frac{3^{-3}m^{5} n^{-1}}{6 }$

This equation can also be written as,

$\frac{m^{5}}{3^{3}6 n^{1} }$

Multiplying the terms in denominator, we have,

$\frac{m^{5} }{162n}$

Thus, the expression which is equivalent to $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ is $\frac{m^{5} }{162n}$

Hence, Option a is the correct answer.

8 0

2 years ago

Read 2 more answers