Find the area of the triangle. Round to the nearest<br> tenth.<br> .

Solve the following system of equations graphically on the set of axes below.

professor190 [17]

The solution of the given equations is: (-1.5, 1.5)

Step-by-step explanation:

Linear equation usually represents a straight line. When a system of two liner equations is given, the point of intersection of lines represented by both equations is the solution of the system of equations.

We will plot the lines of both equations and find their point of intersection (Picture Attached)

We can see in the picture that two lines intersect at (-1.5, 1.5)

Hence,

The solution of the given equations is: (-1.5, 1.5)

Keywords: Linear equations, variables

Learn more about linear equations at:

#LearnwithBrainly

6 0

3 years ago

During a basketball game a player makes 12 of 18 free throws attempted write the ratio as fraction reduced to the lowest term

umka21 [38]

To make this a fraction, let's put $\frac{12}{18}$ . We can divide both top and bottom by 6, leaving us with $\frac{2}{3}$ . This is as far as we can reduce it, making it your answer. I hope this helps!

5 0

3 years ago

Read 2 more answers

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: