Fractions help please??

Dahasolnce [82]

If you want to estimate to the tens, the answer is 680
If you want to estimate to the hundreds, the answer is 700
If you want to estimate to the thousands then the answer would be not an answer
The real answer without estimating is 681

6 0

2 years ago

Read 2 more answers

Consider the matrix shown below:

velikii [3]

Answer:

$A)\ \ \ \ \left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

Step-by-step explanation:

Given the matrix: $\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]$ , it's inverse is calculated using the formula:

$\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]^{-1}=\frac{1}{det\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] }\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]$

#Therefore, we calculate as;

$\frac{1}{det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] }\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\\#det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] =1\\\\\\\\=\frac{1}{1}\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\=\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

Hence, the inverse of the matrix is $\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

8 0

2 years ago

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: