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3 years ago

In a histogram, are the lengths of the rectangles proportional to the width of the bars?

Mathematics

2 answers:

mart [117]3 years ago

8 0

Answer:

Step-by-step explanation:

Length is the frequency density which is obtained by:

Frequency/width

Height is not proportional to width.

Frequency is proportional to the area of the rectangle

Orlov [11]3 years ago

5 0

Answer:

Step-by-step explanation:

In different scenarios, the data will be different. However, sometimes, it's impossible to draw a histogram with equal widths, so in order to maintain clarity and fairness, the area of the bars should actually be proportional to the frequency, which is usually the y-axis of the graph or height of the bars.

Hope this helps!

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3 years ago

Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence Molly's solution is correct.

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence Molly solution is correct.

Learn more about partial product at: brainly.com/question/24716925

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3 years ago

In a histogram, are the lengths of the rectangles proportional to the width of the bars?​

In a histogram, are the lengths of the rectangles proportional to the width of the bars?