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3 years ago

How do i find an I Q R ?

Mathematics

1 answer:

Scrat [10]3 years ago

3 0

Answer:

Sorry if this doesn't help:

Step-by-step explanation:

1, Order the data from least to greatest.

2, Find the median.

3, Calculate the median of both the lower and upper half of the data.

4, The IQR is the difference between the upper and lower medians.

Good luck!

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The box plot represents the number of minutes customers spend on hold when calling a company.

saveliy_v [14]

Answer:

Step-by-step explanation:

By the diagram you can see that the upper quartile is 6. Also this answer was calculated and checked!

4 0

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:

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

4 0

3 years ago

Sequence 120 300 750

likoan [24]

Answer:

2.5

Step-by-step explanation:

This is a geometric sequence that increases by 2.5x.

120 x 2.5 = 300

300 x 2.5 = 750

Hope this helps!

7 0

3 years ago

Please help me. Solve by using substitution.

juin [17]

Answer:

y=-3x+3..........(1)

6x+2y=6

putting value of y

6x+2(-3x+3)=6

6x-6x+6=6

o=0

sorry I do not get answer

6 0

3 years ago

Read 2 more answers

What are the coordinates of A'and B'when AB is reflected in the y-axis? Show your work or explain

vaieri [72.5K]

The coordinates of A'and B'when AB is reflected in the y-axis are; (-2,5) and (-6,3).

<h3>What are the coordinates of A'and B'when AB is reflected in the y-axis?</h3>

According to the task content, it follows that the reflected points A' and B's coordinates are required.

From the graph, the coordinates of A= (2,5) and B= (6,3).

On this note, the coordinates of A' and B' upon reflection over the y-axis is; (-2,5) and (-6,3) respectively.