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3 years ago

Liliana takes 4 breaths per 10 seconds during yoga.At this rate,about how many breaths would liliana take in 2 minutes of yoga

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

4 0

480 accorind to my calcultion

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Ummm yes help me or whatever

Tema [17]

Answer:

i choose whatever

Step-by-step explanation:

yu gave me the option :)

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3 years ago

Simplify (write without the absolute value sign): |x−(−12)|, if x<−12

VladimirAG [237]

Answer:

|x + 12|

Step-by-step explanation:

The question I took form you I assume looks like this...

|x - (-12)|

Therefore, if we simplify this, the answer is |x + 12|

This is the answer because you would multiply -1 by -12

3 0

3 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

padilas [110]

F/34 is the answer. Okokokok

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3 years ago

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

3 years ago

What is the density of 72g over 24cm?

diamong [38]

Mass/volume 72g/24 cubic cm Is equal to 3G per cubic centimeter

5 0

3 years ago