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3 years ago

What is the interquartile range of this data? 6 8 9 20

Mathematics

2 answers:

givi [52]3 years ago

6 0

Answer:

Step-by-step explanation:

The interquartile range is the difference between the values of the upper quartile and the lower quartile.

The upper quartile is the value at the right side of the box, that is 36

The lower quartile is the value at the left side of the box, that is 16

Interquartile range = 36 - 16 = 20

KengaRu [80]3 years ago

5 0

Answer:

Step-by-step explanation:

36 - 16 = 20

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What is the answer?<br><br> Y= -x-4 <br> Y=x

Mashcka [7]

Do you mean like this?

Y = -x - 4
Y = x

x = -x - 4
2x = -4
x = -2

Therefore y = -2

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3 years ago

PLEASEEEEEEEEEEEEEEEEEE URGENT

jeyben [28]

Option C:

f(x) = 5x + 7 is the function of the input-output table.

Solution:

Option A: f(x) = 6x + 9

Input x = 1 in the above equation.

f(1) = 6(1) + 9

= 6 + 9

f(1) = 15

But the output is 12 in the table.

So, it is not the function of the table.

Option B: f(x) = 7x + 5

Input x = 1 in the above equation.

f(1) = 7(1) + 5

f(1) = 12

Input x = 5 in the above equation.

f(5) = 7(5) + 5

f(5) = 40

But the output is 32 in the table.

So, it is not the function of the table.

Option C: f(x) = 5x + 7

Input x = 1 in the above equation.

f(1) = 5(1) + 7

f(1) = 12

Input x = 5 in the above equation.

f(5) = 5(5) + 7

f(5) = 32

Input x = 10 in the above equation.

f(10) = 5(10) + 7

f(10) = 57

Input x = 5 in the above equation.

f(15) = 5(15) + 7

f(15) = 82

All outputs are correct for the given input.

Hence it is the function of the table.

Option D: f(x) = 12x + 1

Input x = 1 in the above equation.

f(1) = 12(1) + 1

f(1) = 12

Input x = 5 in the above equation.

f(5) = 12(5) + 1

f(5) = 61

But the output is 32 in the table.

So, it is not the function of the table.

Hence Option C is the correct answer.

f(x) = 5x + 7 is the function of the input-output table.

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3 years ago

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3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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Answer:

Step-by-step explanation:

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