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Degger [83]
3 years ago
7

Can somebody Answer AND Explain please to be marked brainliest. Click on the photo to see the math problem.

Mathematics
2 answers:
Delvig [45]3 years ago
8 0
The answer is 2.929 unrounded
alexdok [17]3 years ago
6 0
The answer is 4.25
In the first step he never distributed the 1/12 to the 18. He only distributed to the 14
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The table below shows data from a survey about the amount of time students spend doing homework each week. The students were in
nadezda [96]

Answer:

The correct option is;

Both spreads are best described by the standard deviation

Step-by-step explanation:

The given information are;

,                                    College                       High School

High,                              20                               20

Low,                                6                                 3

Q₁,                                   8                                 5.5

Q₃,                                  18                                16

IQR,                                 10                                10.5

Median,                           14                                11

Mean,                              13.3                             11

σ,                                      5.2                             5.4

Checking for outliers, we have

College

Q₁ - 1.5×IQR gives 8 - 1.5×10 = -7

Q₃ + 1.5×IQR gives 18 + 1.5×10 = 33

For high school

Q₁ - 1.5×IQR gives 5.5 - 1.5×10.5 = -10.25

Q₃ + 1.5×IQR gives 16 + 1.5×10.5 = 31.75

Therefore, there are no outliers and the data is representative of the population

From the data, for the college students, it is observed that the difference between the mean, 13.3 and Q₁, 8, and between Q₃, 18 and the mean,13.3 is approximately the standard deviation, σ, 5.2

The difference between the low and the high is also approximately 3 standard deviations

Therefore the college spread is best described by the standard deviation

Similarly for the high school students, the IQR is approximately two standard deviations, the  difference between the mean, 11 and Q₁, 5.5, and between Q₃, 16 and the mean,11 is approximately the standard deviation, σ, 5.4

Therefore the high school spread is also best described by the standard deviation.

4 0
3 years ago
Which change will quadruple the lateral surface area of a cone?
vlada-n [284]
Don't know try something else it will probably help
6 0
3 years ago
Read 2 more answers
Find the unknown. = (65 × 3) + 9
alekssr [168]

Answer:

204

Step-by-step explanation:

Following the order of operations, what is within the parenthesis is solved first.

<em>65 x 3 = 195</em>

Then, you add 9 to 195.

<em>195 + 9 = 204</em>

<em>204</em> is the final answer.

Have a nice day! ^-^

5 0
3 years ago
How many triangles can be constructed with side lengths of 7.2 cm, 6.9 cm, and 12.8 cm?
Alika [10]

Answer:

One triangle

Step-by-step explanation:

-The triangle's dimensions are defined and fixed.

-Given the fixed dimensions, only one unique triangle can be constructed.

-Hence, only one triangle of the dimensions 7.2cm, 6.9cm and 12.8 cm can be formed.

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
2 years ago
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