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3 years ago

Create a set of data with 7 values that has a mean of 30, a median of 26, a range of 50, and an interquartile range of 36.

Mathematics

1 answer:

kotykmax [81]3 years ago

7 0

5, 10, 23, 26, 45, 46, 55

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{c(1) = 3/16<br> {c(n)=c(n−1)⋅4<br> <br><br> what is the 3rd term in this sequence?

mote1985 [20]

Answer:

$\large\boxed{c(3)=3}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}c(1)=\dfrac{3}{16}\\\\c(n)=c(n-1)\cdot 4\end{array}\right$

Put n = 2 and next n = 3 to the recursive formula:

$c(2)=c(2-1)\cdot4=c(1)\cdot4\to c(2)=\dfrac{3}{16}\cdot4=\dfrac{3}{4}\\\\c(3)=c(3-1)\cdot4=c(2)\cdot4\to c(3)=\dfrac{3}{4}\cdot4=3$

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3 years ago

Jayana is trying to run a certain number of miles by the end of the month. If Jayana is 30% of the way to achieving her goal and

torisob [31]

Jayana is trying to run 30 miles by the end of the months.

30% / 100% = 9 / x

9 * 100 = 900

900 / 30 = 30 miles

4 0

3 years ago

The diameter, <img src="https://tex.z-dn.net/?f=D" id="TexFormula1" title="D" alt="D" align="absmiddle" class="latex-formula">,

xenn [34]

Answer:

88.6

Step-by-step explanation:

A=4(pi)r^2/3

D=9.2 : r=4.6

A=4(3.14)(4.6^2)/3=88.5898667

3 0

3 years ago

Find the volume of a rectangular prism with length 20 m,

Cerrena [4.2K]

Answer:

V= 270m³

Step-by-step explanation:

4 0

3 years ago

AYUDA CON ESTO!!! ALGUIEN PORFAVOR

Gre4nikov [31]

Answer:

Problem 1) frequency: 160 heartbeats per minute, period= 0.00625 minutes (or 0.375 seconds)

Problem 2) Runner B has the smallest period

Problem 3) The sound propagates faster via a solid than via air, then the sound of the train will arrive faster via the rails.

Step-by-step explanation:

The frequency of the football player is 160 heartbeats per minute.

The period is (using the equation you showed above):

$Period = \frac{1}{frequency} = \frac{1}{160} \,minutes= 0.00625\,\,minutes = 0.375\,\,seconds$

second problem:

Runner A does 200 loops in 60 minutes so his frequency is:

$\frac{200}{60} = \frac{10}{3} \approx 3.33$ loops per minute

then the period is: 0.3 minutes (does one loop in 0.3 minutes)

the other runner does 200 loops in 65 minutes, so his frequency is:

$\frac{200}{65} = \frac{40}{13} \approx 3.08$ loops per minute

then the period is:

$\frac{13}{40} =0.325\,\,\,minutes$

Therefore runner B has the smaller period

8 0

3 years ago