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3 years ago

How to do n to the power of 6 ÷n to the power of 4

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

5 0

Answer:

n to the power of 2

Step-by-step explanation:

baiscally just subtract 6-4 to get 2

Mashutka [201]3 years ago

3 0

${n}^{6} \div {n}^{4} \\ = {n}^{(6 - 4)} \\ = {n}^{2}$

This is the answer.

Hope it helps!!

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The value of 2^3 + 4^2 = ___. Numerical Answers Expected! Answer for Blank 1:

vekshin1

Answer:

Step-by-step explanation:

2^3+4^2

2(2)2+4(4)

8+16

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3 years ago

Two congruent angles have the same angle measure.

muminat

Answer:

True is the ans.

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Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: