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2 years ago

PLEASEEE HELP! What is the simplest radical form of the expression? (8x^7y^4)^2/3

Mathematics

1 answer:

Leona [35]2 years ago

5 0

Answer:

4x^4 y^2 ∛x^2y^2

Step-by-step explanation:

$(8x^7y^4)^\frac{2}{3} \\\\8^\frac{2}{3}(x^7)^\frac{2}{3}(y^4)^\frac{2}{3}$

$(2^3)^\frac{2}{3}x^7*^\frac{2}{3}y^4*^\frac{2}{3}\\4x^\frac{14}{3}y^\frac{8}{3}$

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Multiply 5/12 and the multiplicative inverse of -1/4

IgorC [24]

Answer:

The answer is -2

Step-by-step explanation:

Hope it helps :)

4 0

3 years ago

What is 1/5 + 1/2 in a fraction<br>

Bezzdna [24]

Answer:

7/10

Step-by-step explanation:

first you find a common denominator which would be 10 so you turn 1/5 into 2/10 and 1/2 into 5/10 and then add 2/10 + 5/10 =7/10

4 0

3 years ago

Read 2 more answers

If f(x) = 4x2 - 5x + 7 and g(x) = 3x2 - 2x + 8, find f(x) + g(x). A) 7x2 - 7x + 15 B) 7x4 - 7x2 + 15 C) 7x2 + 15 D) 12x4 - 8x3 +

lutik1710 [3]

Answer:

A) 7x^2-7x+15

Step-by-step explanation:

f(x) = 4x2 - 5x + 7,

g(x) = 3x2 - 2x + 8

f(x) +g(x) =

4x^2 - 5x + 7 + 3x^2 - 2x + 8=

7x^2-7x+15

8 0

3 years ago

The probability that aperson living in a certaincity owns a dog is estimated to be 0.3. Find the probabilitythat the tenth perso

allochka39001 [22]

The probability that the tenth person randomly interviewed in that city is the ﬁfth one to own a dog = 0.051

Given,

Consider the interview of a person is a trial.

Lets consider it success if a person owns a dog and consequently , a person without a dog will be a failure.'

Let the random variable X represent the number of persons to be interviewed to get 5 dog owner: in other words, to get 5 successors.

Probability of a success in each trial is p = 0.3. Therefore probability of a failure in each trial is q = 1 - p = 0.7 .

Because trials are independent. X has a negative binomial distribution with parameters k = 5, p = 0.3

P(X =x ) = b(x: k, p) = $[(x-1)(k-1)] p^kq^r^-^k$

where, x = k , k + 1 , k + 2......(1)

Now, Lets find the probability that the tenth person randomly interviewed in that city is the fifth one to own a dog

Using equation (1), we get:

P(X = 10) = b(10: 5, 0.3)

= [(10-1) (5 - 1)] $(0.3)^5(0.7)^1^0^-^5$

= [(9)(4)] $(0.3)^5(0.7)^1^0^-^5$

= 0.05146

Hence, The probability that the tenth person randomly interviewed in that city is the ﬁfth one to own a dog = 0.051

Learn more about Probability at:

brainly.com/question/743546

#SPJ9

4 0

1 year ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago