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3 years ago

What are 5 ways that graph can be misleading ?

Mathematics

2 answers:

Elza [17]3 years ago

3 0

<h3>Hello there hope you having a good day :)</h3><h3>. The axis are plotted to a scale that make people think there is more or less value.</h3><h3>. The graph overall may represent a rather insignificant subject.</h3><h3>. The graph overall may make something important appear rather small.</h3><h3>. The graph may represent factors without pointing out other important factors.</h3><h3>. If there is no x and y axis.</h3><h3>Hopefully this helps you. Another one would be if it doesn't start with 0.</h3>

Nata [24]3 years ago

3 0

There are many ways that graphs can be misleading. Depending on how you set up your graph, the same set of data can look very different. It's important to understand that there are some common mistakes to avoid when making graphs. In addition, advertisers have been known to intentionally make their graphs misleading to make their data match their point.

In the image provided, this is a good example of a misleading graph.

First, notice that we know that the first bar is 10 and the second bar is 50. However, there are no labels on the horizontal or vertical axis so we don't know what the bars represent.

In addition, notice that there is no title on this graph. Since there is no title on the graph, we don't know anything about the data.

Further, look at the scale on the vertical axis. Notice that it's an uneven scale. The uneven scale makes it appear that the first bar is closer to the second bar than it actually is.

We could correct this graph by adding labels to the x and y-axis, adding a title to the graph, and making the scale on the vertical axis even.

Another way graphs can be misleading is using the wrong type to represent data. For example, a scatter plot might be better than a regular graph in some cases.

Lastly, graphs can be misleading by adding a wide range of numbers to show and increase or decrease relatively small to how it actually is.

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10. multiply 4x(x3 2)

DedPeter [7]

It is 10x4x3x2 equals two hundred and forty

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3 years ago

Can someone please help me

jeka94

Answer:

A. 1200

Step-by-step explanation:

Speed in meter per minute

= (216*1000)/(3*60)

= 216000/180

= 1200

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2 years ago

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The slope f′(x) at each point (x,y) on a curve y=f(x) is given, along with a point (a,b) on the curve. Use this information to f

Montano1993 [528]

$f'(x)=\dfrac{4x}{1+7x^2}$

Integrating gives

$f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx$

To compute the integral, substitute $u=1+7x^2$ , so that $\frac27\,\mathrm du=4x\,\mathrm dx$ . Then

$f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C$

Since $u=1+7x^2>0$ for all $x$ , we can drop the absolute value, so we end up with

$f(x)=\dfrac27\ln(1+7x^2)+C$

Given that $f(0)=10$ , we have

$10=\dfrac27\ln1+C\implies C=10$

so that

$\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}$

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3 years ago

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

$P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 18 Southwest flights

r = number of success = at least 13 flights arrive late

p = probability of success which in our question is probability that

flights arrive late, i.e. p = 1 - 0.80 = 20%

Let X = <u><em>Number of flights that arrive late</em></u>.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X $\geq$ 13)

P(X $\geq$ 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= $\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}$

= $\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}$