All categories

3 years ago

When (81x^2/7) (2/9x^3/7) is simplified, it can be written in the form ax^b where a and b are real numbers. Find ab.

Mathematics

1 answer:

larisa86 [58]3 years ago

5 0

Answer:

In improper form your solution will be $\frac{90}{7}$ . As a mixed fraction it will be $12\frac{6}{7}$ .

Step-by-step explanation:

The first thing we want to do here is to simplify this expression. After doing so, " a " and " b " should be multiplied to result in a possible improper fraction,

$\left(81x^{\frac{2}{7}}\right)\:\left(2/9x^{\frac{3}{7}}\right)\:$ - Apply exponential rule " $\:a^b\cdot \:a^c=a^{b+c}$ "

= $81\cdot \frac{2}{9}x^{\frac{2}{7}+\frac{3}{7}}$ - Combine fractions $\frac{2}{7}$ and $\frac{3}{7}$

= $81\cdot \frac{2}{9}x^{\frac{5}{7}}$ - Multiply the fractions, and simplify further

= $\frac{162x^{\frac{5}{7}}}{9}$ = $18x^{\frac{5}{7}}$ - This is out simplified expression

Now that we have this simplified expression, we can see that a = $18$ , and b = $\frac{5}{7}$ . Therefore, multiplying the two we should receive the improper fraction as follows,

$18 * \frac{5}{7}$ = $\frac{90}{7}$ - Note that this is in improper form. If you want your solution in a mixed fraction, it will be $12\frac{6}{7}$ .

You might be interested in

slader The magnitude of earthquakes recorded in a region of North America can be modeled ashaving an exponential distribution wi

Akimi4 [234]

Answer:

a) $P(X>3.0)=1-P(X \leq 3.0)= 1- [1- e^{-\frac{1}{2.4} 3.0}]=e^{-\frac{1}{2.4} 3.0}=0.287$

b) $P(2$

$P(2$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

$P(X>x)= e^{-\lambda x}$

We can express in terms of the mean $\mu =\frac{1}{\lambda}$

$P(X>x)= e^{-\frac{1}{\mu} x}$

And the cumulative function would be given by the complement rule like this:

$P(X\leq x)=1- e^{-\frac{1}{\mu} x}$

Solution for the problem

For this case we have that X the random variable that represent the magnitude of earthquakes recorded in a region of North America, we know that the distribution is given by:

$X\sim Expon(\mu =2.4)$

a) Find theprobability that an earthquake striking this region will a exceed 3.0 on the Richter scale

So for this case we want this probability:

$P(X>3.0)=1-P(X \leq 3.0)= 1- [1- e^{-\frac{1}{2.4} 3.0}]=e^{-\frac{1}{2.4} 3.0}=0.287$

b) fall between 2.0 and 3.0 on the Richter scale.

For this case we want this probability:

$P(2 < X$

And replacing we have this:

$P(2$

7 0

3 years ago

Factor completely 2x^3-18x

Alexxandr [17]

Answer:

2x(x+3)(x-3)

Step-by-step explanation:

3 0

3 years ago

Kevin installed a certain brand of automatic garage door opener that utilizes a transmitter control with five independent switc

aleksandrvk [35]

Answer:

0.03125

Step-by-step explanation:

Number of Combinations possible with the Transmitters = 2*2*2*2*2 = 32

Probability that they will use the same settings = 1 - Probability of them using different settings.

= 1 - 32*31/(32^2)

=1 - 992/1024

= 1 - 0.96875

= 0.03125

4 0

2 years ago

What number must be added to14,056 to result the sum of 38,776

Soloha48 [4]

You must add 24,720

Just do 38,776 minus 14,056 when you need those types of answers.

8 0

2 years ago

20 different comic books will be distributed to five kids. (a) How many ways are there to distribute the comic books if there ar

sveta [45]

Answer:

a) 21684

b) 15504

Step-by-step explanation:

a) Distribute the comic books as follows

only to 1 kid one way but we have five kids then 5

to two kids C₂₀,₂ = 20! /2! ( 20-2)! = 20*19 /2 = 190

To three C₂₀,₃ = 20! /3! ( 20 - 3)! = 20*19*18 /6 = 1140

To four C₂₀,₄ = 20 ! / 4! ( 20 - 4 )! = 20*19*18*17 /4*3*2 = 4845

To five C₂₀,₅ = 20! /5! (20 - 5 )! = 20*19*18*17*16/5*4*3*2*1

C₂₀,₅ = 15504

Then total ways of distribution are:

5 + 190 + 1140 + 4845 + 15504 = 21684

b) C₂₀,₅ we know from a that is equal to 15504

7 0

2 years ago