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3 years ago

How can I get the median when the amount of number are 8???

Mathematics

2 answers:

alina1380 [7]3 years ago

8 0

You can take the average of the two middle numbers

Dafna11 [192]3 years ago

7 0

Then, the median is found by taking the mean (or average) of the two middle most numbers.

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X/4 + 9 = -13 help please

Murljashka [212]

Answer:Subtract

from both sides of the equation

from both sides of the equation

Multiply all terms by the same value to eliminate fraction denominators

7 0

2 years ago

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Can you explain to me (detailed) on how to do this problem? I will give you a brainly if you get this right.

Pepsi [2]

Answer:

9 cm

Step-by-step explanation:

The formula to find the area of a square is Base times height (B x H)

Since a square has equal sides, it means each side is 3 centimeters.

So, all you have to do is 3 times 3.

3 x 3=9

<u>Final answer</u>

9 cm

4 0

3 years ago

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Can someone please help me with these 7 questions please?

yarga [219]

$(1)\ (-xy)^3(xz)$

Expand

$(-xy)^3(xz) = (-x)^3* y^3*(xz)$

$(-xy)^3(xz) = -x^3* y^3*xz$

Rewrite as:

$(-xy)^3(xz) = -x^3*x* y^3*z$

Apply law of indices

$(-xy)^3(xz) = -x^4y^3z$

$(2)\ (\frac{1}{3}mn^{-4})^2$

Expand

$(\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}$

$(\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8$

$(3)\ (\frac{1}{5x^4})^{-2}$

Apply negative power rule of indices

$(\frac{1}{5x^4})^{-2}= (5x^4)^2$

Expand

$(\frac{1}{5x^4})^{-2}= 5^2x^{4*2}$

$(\frac{1}{5x^4})^{-2}= 25x^{8$

$(4)\ -x(2x^2 - 4x) - 6x^2$

Expand

$-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2$

Evaluate like terms

$-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2$

Factor out x^2

$-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2$

Factor out -2

$-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2$

$(5)\ \sqrt{\frac{4y}{3y^2}}$

Divide by y

$\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}$

Split

$\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}$

Rationalize

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}$

$(6)\ \frac{8}{3 + \sqrt 3}$

Rationalize

$\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}$

Apply different of two squares to the denominator

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}$

Simplify

$\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}$

$(7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}$

Expand

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}$

Split

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}$

Evaluate all roots

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}$

$(8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}$

Expand

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}$

Factorize each

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}$

Factor out (r+3) in the numerator and (r + 6) in the denominator

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}$

Cancel out r - 2

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}$

$(9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}$

Cancel out x

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}$

Expand the numerator of the 2nd fraction

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}$

Factorize

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}$

Factor out x - 7

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Factor out 4 from 4x + 8

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Cancel out x + 2

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}$

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}$

$(10)\ (3x^3 + 15x^2 -21x) \div 3x$

Factorize

$(3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x$

Cancel out 3x

$(3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7$

$(11)\ \frac{m}{6m + 6} - \frac{1}{m+1}$

Take LCM

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}$

Expand

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}$

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}$

$(12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}$

Rewrite as:

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}$

Express as multiplication

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}$

Express y^2 - 9 as y^2 - 3^2

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}$

Express as difference of two squares

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}$

Read more at:

brainly.com/question/4372544

3 0

2 years ago

Find Y. <br> Round to the nearest tenth.<br> Y= ? Degrees

lana66690 [7]

Answer:

Y = 59.0°

Step-by-step explanation:

Using the cosine rule in terms of x, y and z

cosY = $\frac{x^2+z^2-y^2}{2xz}$

with x = 22, y = 20 and z = 18

cosY = $\frac{22^2+18^2-20^2}{2(22)(18)}$

= $\frac{484+324-400}{792}$

= $\frac{408}{792}$ , thus

Y = $cos^{-1}$ ( $\frac{408}{792}$ ) ≈ 59.0° ( to the nearest tenth )

4 0

3 years ago

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A tennis player makes a successful first serve 51% of the time. If she serves 9 times, what is the probability that she gets exa

aleksley [76]

Answer:

$P(X=3)$

And we can use the probability mas function and we got:

$P(X=3)=(9C3)(0.51)^3 (1-0.51)^{9-3}=0.1542$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=9, p=0.51)$