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2 years ago

If the outliers are not included what is the mean of the data set 76,79,80,82,50,78,79,81,82

Mathematics

2 answers:

iVinArrow [24]2 years ago

8 0

Answer:

Step-by-step explanation:

mean=[76+ 79 + 80 + 82+ 78 + 83 + 79 + 81 + 82]/9

mean=[720]/9

mean=80

defon2 years ago

3 0

Answer:

The answer is 80

Step-by-step explanation:

we know that

the outlier is 50, as it is not around the other numbers in the data set.

therefore

mean=[76+ 79 + 80 + 82+ 78 + 83 + 79 + 81 + 82]/9

mean=[720]/9

mean=80

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Solve for x: 3 over 3x+ 1 over x+4= 10 over 7x

hram777 [196]

Answer:

Step-by-step explanation:

3/3x + 1/x+4 = 10/7x

3/3x + 1/x+4 = 10/7x , x=0,x=-4

1/x + 1/x+4 = 10/7x

1/x + 1/x+4 - 10/7x = 0

7(x+4)+7x-10(x+4) / 7x*(x+4) = 0

7x+28+7x-10x-40 / 7x*(x+4) = 0

4x-12 / 7x*(x+4) = 0

4x-12=0

4x-=12

x=3,x=0,x=-4

x=3

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3 years ago

A jeweler orders necklaces from a website that offers $6 shipping for any-size order. Each necklace costs $7. The jeweler wants

Furkat [3]

Answer:

c = $7n + $6

Step-by-step explanation:

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3 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

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

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago