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3 years ago

Plzzzz help me!!

Mathematics

2 answers:

Airida [17]3 years ago

6 0

let

x---------> the length side of the box

y---------> the width side of the box

we know that

h=3 in

V=390 in³

volume of the box=x*y*3-------> 390=x*y*3-----> x*y=130------> equation 1

x=3+y------> equation 2

substitute 2 in 1

[3+y]*y=130-----> y²+3y=130------> y²+3y-130=0

using a graph tool------> to resolve the second order equation

see the attached figure

the solution is

y=10 in

x=3+y-----> x=3+10-----> x=13 in

the answer is

the width is 10 in

Vlad1618 [11]3 years ago

6 0

Answer:

Volume is length times width times height. So, you would need to do 9 times 3 (27) times 10 (270). Then, you would divide by 10, because it is scaled down by 1/10, leaving you with 27

Step-by-step explanation:

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Read 2 more answers

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

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Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

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4 years ago