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3 years ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

Mumz [18]3 years ago

5 0

It’s b i believe, if i did the math correctly

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Which one i greater 2/8 or 1/2 ?

ryzh [129]

$\frac{2}{8}=\frac{1}{4}\\
\frac{1}{2}=\frac{2}{4}\\\\
\frac{2}{4}>\frac{1}{4} \Rightarrow\frac{1}{2}>\frac{2}{8}$

5 0

3 years ago

How do I do this?<br>*Look at the directions in the photo*

lora16 [44]

Answer:

$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

Step-by-step explanation:

$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

$For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2$

$For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2$

$Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2$

7 0

3 years ago

HELP ASAP PLSSSSSSSSSS

inysia [295]

Answer:

Neighborhood Q appears to have a bigger family size

Step-by-step explanation:

Mean = the sum of all data values divided by the total number of data values

Number of families in Neighborhood Q = 9

Mean family size of Neighborhood Q:

= (2 + 5 + 4 + 3 + 2 + 5 + 3 + 6 + 5) ÷ 9

= 35 ÷ 9

= 3.888888...

Number of families in Neighborhood S = 9

Mean family size of Neighborhood S:

= (2 + 3 + 2 + 3 + 7 + 2 + 3 + 3 + 2) ÷ 9

= 27 ÷ 9

= 3

The mean family size of Neighborhood Q is 3.88.. and the mean family size of Neighborhood S is 3. Therefore, Neighborhood Q appears to have a bigger family size as it's average family size is bigger than that of Neighborhood S.

6 0

2 years ago

A computer manufacture a quality check to make sure the computer sound 1004 head what is experimental probability of the compute

FinnZ [79.3K]

Answer:y

Step-by-step explanation:

7 0

3 years ago

I AM GIVING BRAINLIEST

spayn [35]

Answer:

The first choice is correct

Step-by-step explanation:

$32.00 x 15% = 4.80. 32.00-4.80=27.20. Another way to work it is the first choice, subtracting 15% from 100% and then multiplying 85% by $32.00. Same answer

7 0

3 years ago