All categories

3 years ago

How can you use bar graphs to compare quantities in real life situations

Mathematics

2 answers:

SashulF [63]3 years ago

7 0

We can use bar graphs to solve many real life problems. A bar graph is simply a graphical display of data using bars of different heights. To answer this question let's take an example. Suppose you are a student and want to know which kind of food people prefer in your urban area. So, you did a survey to find the answer. In this way, you found that:

40 People prefer Hamburgers
50 People prefer Hot Dogs
60 People prefer Fried Chicken
10 People prefer Roast Beef

So, we can show the relative sizes using a bar graph. This is indicated in the Figure below. Therefore, you can see that most people prefer Hamburgers and few people prefer Roast Beef

BlackZzzverrR [31]3 years ago

4 0

You can illustrate it by comparing two quantites in a correlational basis. Like the price times the time of how a courier delivers a package or a internet speed vs the data cap and land area and the investment.

Bar graph display directly the variables which are the rate and ratio of the numbers to visualize and display the results, these contrasting the different outcomes. On the contrary, histogram is used in grouped frequency parameters. Moreover, as little as the given five parameters or data set this will be ineffective and will result to a bar graph only and basically, the suited option is the aforementioned vertical graph to display the numbers. To expound on the definition of histogram it is used when the frequency is grouped. For example the data set of 1-5, 6-10, 11-15 and 16-20 this now can be used and applied to illustrate histogram because of the number and quantity of the given data.

You might be interested in

HELP!! Complete the ratio table. 13 15 26 _ 45 _ 60 _ 65 75

sladkih [1.3K]

Answer:

30,52,69

Step-by-step explanation:

8 0

2 years ago

Tim has 80 coins in his piggy bank. Of these coins, 28 are nickels and the rest are quarters. What percent of the coins in the b

KIM [24]

65 percent I check it

8 0

2 years ago

Read 2 more answers

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using integration by parts:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

1 year ago

Can you conclude that the parallelogram is a rhombus, a rectangle, or a square? Explain. PLEASE HELP! WILL MARK BRAINLIEST!!!!

ludmilkaskok [199]

Answer: A parallelogram is a rectangle.

Step-by-step explanation:

This is always true. Â Squares are quadrilaterals with 4 congruent sides and 4 right angles, and they also have two sets of parallel sides. Parallelograms are quadrilaterals with two sets of parallel sides. ... A parallelogram is a rectangle.

5 0

1 year ago

Luke scored 21 goals in 7 games. if he scores goals at the same rate how many goals will he score in 23 games

Gre4nikov [31]

Luke scored 21 goals in 7 games. if he scores goals at the same rate how many goals will he score in 23 games
the answer is 69
you divide 21 by 7 which is 3 and you multiply 3 by 23 and you get 69

3 0

3 years ago