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AleksAgata [21]
3 years ago
14

A rectangular box is 6 inches wide 10 inches long and 2 inches tall how much wrapping paper is needed to cover the box exactly

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

184 in²

Step-by-step explanation:

Given :

Width, w = 6 inches

Length, l = 10 inches

Height, h = 2 inches

To obtain how much wrapping paper is needed ; we take the surface area of the box

Surface area = 2(lw + lh + wh)

Surface area = 2((6*10) + (6*2) + (10*2))

Surface area = 2(60 + 12 + 20)

Surface area = 2(92)

Surface area = 184 in²

The amount of wrapping paper needed = 184 in²

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How does the speed of a runner vary over the course of a marathon (a distance of 42.195 km)? Consider determining both the time
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The observed typical difference value of mode is 100 seconds approximately.

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

Step-by-step explanation:

Fundamentals

Consider that there are x favorable cases to an event E, out of a total of n cases. Then, the probability of that event is written as:

P(E)=  

Total number of cases /Number of favorable cases =  x/n

A histogram is constructed for continuous data, which is divided into classes called bins. The shape of the distribution can be determined from the histogram.

step 1

The provided histogram indicates time difference on xx -axis and frequency of runners on yy -axis. To determine the typical difference value, identify the peaks of the graph.

The histogram is skewed towards right side. The graph indicates that there are few outliers around 700 seconds. For a typical difference value, the value of mode is considered.

The graph indicates that the value of mode is 100 seconds approximately.

The observed typical difference value of mode is 100 seconds approximately.

Explanation

From the histogram, the typical difference value is obtained on the basis of guessing the value of mode which has been approximated to be around 100 seconds.

Step 2

From the histogram, it can be estimated that there are around 10 runners which has negative difference which means approxi8matley 10 runners ran the late distance more quickly than the early distance,

The approximate sample size can be calculated as:

Sample size=90+190+180+160+120+80+60+40+30+20

Thus, the proportion of runners is obtained as:

\begin{array}{c}\\p = \frac{{\left( \begin{array}{l}\\{\rm{Number of runners who has }}\\\\{\rm{negative time difference}}\\\end{array} \right)}}{{{\rm{Total sample size}}}}\\\\ = \frac{{10}}{{970}}\\\\ = 0.01\\\end{array}  

p=  <u> </u>Number of runners who has

<u>     negative time difference   </u>

      Total sample size

​  

=10/970

=0.001

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

EXPLANATION

The obtained proportion is 0.01. It indicates that there are approximately 1% of the runners who ran late distance more quickly than the early distance is very few.

​  

​  

3 0
3 years ago
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