Write 31 1/4% as a fraction

vivado [14]

Answer:

Any number which is on left side on the line number is less than the number which is on right hand side relative to that number.

As -28 is on the right hand side of the line number, and -162 is heading towards left hand side relative to -28.

Therefore, -28 is greater than -162.

Step-by-step explanation:

To determine:

Is -28 greater than or less than -162?

Solution Steps:

In mathematics, a number line is considered to be a straight line with numbers placed at equal intervals along its length.

A number line could be extended infinitely in any direction i..e -∞ to +∞.

A number is usually represented horizontally.

Any number which is on left side on the line number is less than the number which is on right hand side relative to that number.

As -28 is on the right hand side of the line number, and -162 is heading towards left hand side relative to -28.

Therefore, -28 is greater than -162.

In other words,

$-28>-162\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:\mathrm{True\:for\:all}\:x\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}$

Keywords: number line, less than, greater than

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7 0

3 years ago

PLEASE HELP! 50 POINTS! (Solve each system of equations. Determine which quadrant the solution lies in.)

Sloan [31]

Answer:

wet

Step-by-step explanation:

3 0

2 years ago

AYOOO PLZ HELP ASAP!!!

tamaranim1 [39]

Answer:

B.

Step-by-step explanation:

Well we know that

$224=2^{5} *7$

so we can get the 2 outside of the radical

$x^{11} =(x^{5} )^{2} *x$

and we can get the x^2 outside too.

$y^8=y^5*y^3$

and we also can get y outside.

so we have:

$2x^{2}y\sqrt[5]{7xy^3}$

7 0

3 years ago

2. Using the substitution method, Jose is solving the following system of equations

Svetradugi [14.3K]

Answer:

3

Step-by-step explanation:

the answer is 3 if you check very well

5 0

3 years ago

Find the standard deviation of the following data. Round your answer to one decimal place. x 0 1 2 3 4 5 P(X

lbvjy [14]

Answer:

$\sigma = 1.8$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} \ \\ P(x) & {0.2} & {0.1} & {0.1} & {0.2} & {0.2}& {0.2} \ \end{array}$

Required

The standard deviation

First, calculate the expected value E(x)

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * 0.2 + 1 * 0.1 + 2 * 0.1 + 3 * 0.2 + 4 * 0.2 + 5 * 0.2$

$E(x) = 2.7$

Next, calculate E(x^2)

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * 0.2 + 1^2 * 0.1 + 2^2 * 0.1 + 3^2 * 0.2 + 4^2 * 0.2 + 5^2 * 0.2$

$E(x^2) = 10.5$

The standard deviation is:

$\sigma = \sqrt{E(x^2) - (E(x))^2}$

$\sigma = \sqrt{10.5 - 2.7^2}$

$\sigma = \sqrt{10.5 - 7.29}$

$\sigma = \sqrt{3.21}$

$\sigma = 1.8$ --- approximated

5 0

3 years ago