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3 years ago

What 2 things are recommended in order for you to have a neat graph

Mathematics

1 answer:

Oksana_A [137]3 years ago

6 0

Answer:

use a ruler

buy a graph

Step-by-step explanation:

Not the best answer tho... becuase this useless Mod deleted all my answers randomly

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What's the equation of a line that passes through points (–1, –5) and (1, 3)?

erastova [34]

Answer:

y=4x-1

Step-by-step explanation:

4 0

3 years ago

Jason inherited a piece of land from his great-uncle. Owners in the area claim that there is a 45% chance that the land has oil.

Licemer1 [7]

So,

We are trying to find the compound probability of there BEING oil and the test predicting NO oil.

The percent chance of there actually being oil is 45%. We can convert this into fraction form and simplify it.

45% --> $\frac{45}{100}$

$\frac{45}{100}--\ \textgreater \ \frac{3*3*5}{2*2*5*5}$

$\frac{3*3*5}{2*2*5*5}--\ \textgreater \ \frac{3*3}{2*2*5}$

$\frac{3*3}{2*2*5}--\ \textgreater \ \frac{9}{20}$

That is the simplified fraction form.

The kit has an 80% accuracy rate. Since we are assuming that the land has oil, we need the probability that the kit predicts no oil.

The probability that the kit detects no oil will be the chance that the kit is not accurate, which is 20% (100 - 80 = 20). We can also convert this into fraction form and simplify it.

20% --> $\frac{20}{100}$

$\frac{20}{100}--\ \textgreater \ \frac{2}{10}$

$\frac{2}{10}--\ \textgreater \ \frac{1}{5}$

That is the probability of the kit not being accurate (not predicting any oil).

To find the compound probability of there being oil and the kit not predicting any oil, we simply multiply both fractions together.

$\frac{9}{20}*\frac{1}{5}$

$\frac{9}{20*5}$

$\frac{9}{100}$

So the probability of there BEING oil and the kit predicting NO oil is 9 in 100 chances.

3 0

3 years ago

Read 2 more answers

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

3 years ago

What is the answer to this question

kobusy [5.1K]

Answer:

The 3rd and 4th choices are correct.

Step-by-step explanation:

The x values and y values are reversed from each other.

6 0

3 years ago

(25 POINTS!!)What is the equation of the line that passes through the points (15,9) and (-2,9)?

Viktor [21]

Let's use the point-slope formula...

y-y1=m(x-x1)

m is the slope.

y--9=7(x--2)

Subtracting a negative number is the same as adding a positive number...

y+9=7(x+2)

y+9=7x+14

Let's subtract 9 from both sides...

-9+y+9=7x+14-9

y=7x+5

The formula is now in the format of...

y=mx+b

This is known as slope-intercept.

m is the slope.

b is the y-intercept, the value of y when x=0.

Standard formula is...

Ax+By=C

Neither A nor B equal zero.

A is greater than zero.

y=7x+5

Let's move 7x to the left side of the equation. It becomes negative.

-7x+y=5

Let's multiply both sides by -1 to render A greater than zero.

-1(-7x+y)=(5)(-1)

7x-y=-5 

This is the equation in standard form.

5 0

4 years ago

Read 2 more answers