How do I make a graph with y axis and x axis using numbers like 180

Mathematics

1 answer:

Rashid [163]3 years ago

8 0

Generally, you choose some convenient number for each small division of the graph to represent. Usually, 1, 2, 5, 10, and multiples of 10 times those are considered "convenient."

Here, it might do to use 5 units for each small division, so that 180 is 36 small divisions. If you have 10 small divisions per inch of your graph, then your graph will be about 4 inches high (if you start from zero).

Since the range of your graph is from 112 to 180, about 70 units, you could actually use 1 unit for each small division. Again, if you are using 10 divisions per inch, your graph will be about 7 inches high, so will fit nicely on a page. In this case, the lower edge will start at 110, not zero.

Since this is a temperature plot, you expect the curve to be exponential, with a lower limit of room temperature. Thus, you may want to use a graph with a range that includes room temperature at the low end. That will affect your choice of vertical scale somewhat.

You might be interested in

Is f(x)=-3 an odd function?

Ann [662]

Its an even function

4 0

3 years ago

Todor was trying to factor 10x^2-5x+15

valkas [14]

Answer: 2x^2-x+3

Step-by-step explanation:

7 0

3 years ago

Negative -5/8 - 1/2= <br> I wanna know fast anyone??

alexira [117]

Answer" -9/8

by-step explanation:

-5/8 - 1/2 = -9/8

Simplify: -9/8= 1 1/8

3 0

3 years ago

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: