All categories

2 years ago

What is an statistics

Mathematics

1 answer:

Vlad [161]2 years ago

7 0

Hello There!

Statistics is a tool used to help people process, summarize, analyze and interpret data for making better decisions.

It helps us with future events if we know what has happened in past.

Helps us make better decisions.

The better we know about the future of an outcome, the more possibility we have to work around it.

You might be interested in

Domain and range of linear and quadratic functions worksheet

Vlad [161]

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

<h3>What is a quadratic equation?</h3>

It's a polynomial with a worth of nothing.

There exist polynomials of variable power 2, 1, and 0 terms.

A quadratic condition is a condition with one explanation where the degree of the equation is 2.

Domain and range of linear and quadratic functions

Let the linear equation be y = mx + c.

Then the domain and the range of the linear function are always real.

Let the quadratic equation will be in vertex form.

y = a(x - h)² + k

Then the domain of the quadratic function will be real.

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

More about the quadratic equation link is given below.

brainly.com/question/2263981

#SPJ4

7 0

1 year ago

5-4. Solve each of the following equations for the indicated variable. Use your Equation Mat if it is helpful. Write down each o

balu736 [363]

QUESTION 1

The given equation is :

$2(y - 3) = 4$
We want to solve for y. This means we have to isolate y on one side of the equation while all other constants are also on the other side of the equation,

We first divide both sides by 2 to obtain,

$\frac{2(y - 3)}{2} = \frac{4}{2}$

We cancel out the common factors to get,

$y - 3 = 2$

We now group like terms to get,

$y = 3 + 2$

We simplify to obtain,

$y = 5$

QUESTION 2

The given equation is

$2x + 5y = 10$
We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

This implies that,

$2x= 10 - 5y$

We now divide through by 2, to obtain,

$\frac{2x}{2} = \frac{10}{2} - \frac{5y}{2}$

This will now give us,

$x = 5 - \frac{5}{2} y$

QUESTION 3

The given equation is
$6x + 3y = 4y + 11$

We want to solve for y. This means that we need to isolate y on one side of the equation, while all other variables and constants are also on the other side of the equation.

We first of all group all the y terms on one side of the equation to obtain,

$6x - 11 = 4y - 3y$

We simplify to get,

$6x - 11 = y$

This implies that,

$y = 6x - 11$

QUESTION 4

The given equation is,

$3(2x + 4) = 2 + 6x + 10$

We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

Let us first expand the brackets to get,

$6x + 12 = 2 + 6x + 10$
This implies that

$6x + 12 = 6x + 12$

We group like terms to get,

$6x - 6x = 12 - 12$

We now simplify both sides to get,

$0x = 0$

We don't want x to vanish, so let us try to divide both sides by zero to get,

$x = \frac{0}{0}$

This is an indeterminate form, which implies that, x has infinitely many solutions. This means that all the real numbers are solution to the equation.

$\therefore \: x \in \: R$

QUESTION 5

The given equation is,

$y = - 3x + 6$

We want to solve for x, so we add the additive inverse of 6, which is -6 to both sides of the equation to get,

$y - 6 = - 3x$

We rewrite this to obtain,

$- 3x = y - 6$

We divide through by -3 to get,

$\frac{ - 3x}{ - 3} = \frac{y}{ - 3} - \frac{6}{ - 3}$

This will give us,

$x = - \frac{y}{3} + 2$

or

$x= 2 - \frac{1}{3} y$