Simplify to a single power of 5: (5^4)^3

Mathematics

1 answer:

Vlada [557]2 years ago

5 0

Answer:

5^12

Step-by-step explanation:

This is called the "Power of Powers rule." When you are doing this rule, the inside exponent inside the parenthesis (the ^4 in this case) multiplies with the one outside the parenthesis, like this: 5^4*3 = 12.

However, if there is a base number next to the exponent <em>outside</em> the parenthesis that is a <em>different</em> number than the one inside the parenthesis, this method will not work. (Example: (5^4)5^3)

Important note: Anytime you are combining exponents through any rule, the base number must be the same to be able to combine.

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Offering 20 Points(not a lot I know but I really need quick help)!!!

myrzilka [38]

Answer:

Correct arrangement of equation of displacement to find a is as follows;

1- Vt - d = 1/2 a t^2 (^ represents exponent i.e. t square as given in equation)

2- 2(Vt - d ) = a t^2

3- a = 2(Vt - d )/ t^2 (keep in mind, 2(Vt - d) whole divided by t^2)

Step-by-step explanation:

1- In the first equation, Vt is taken to the left side of the equation (keep in mind, original equation of displacement used for reference as given in question) and multiplied by -1 on the both sides of the equation.

2- In the second equation, 2 is multiplied on the both sides.

3- Multiply t^2 on both sides of the equation, We will get a in correct arrangement, which is required to find.

5 0

3 years ago

"is it appropriate to use the normal approximation for the sampling distribution of"

Katyanochek1 [597]

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution $B(n,p)$ , where $n$ is the number of trials and $p$ is the probability of success. It can be approximated by the Normal distribution with the mean of $np$ and the variance of $np(1-p)$ , denoted by $N(np,np(1-p))$ if the following condition is met:

$n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)$

2. Poisson distribution: Let the sampling distribution be the Poisson distribution $P(\lambda)$ where $\lambda$ is its mean. It can be approximated by the Normal distribution with the mean $\lambda$ and the variance $\lambda$ , denoted by $N(\lambda,\lambda)$ when $\lambda$ is large enough, say $\lambda>1000$ (however, different sources may give different lower value for $\lambda$ but the greater it is, the better the approximation).