Ask question

Ask question

All categories

3 years ago

14

Jackie cut a piece of paper along its diagonal, as shown below, forming two triangles. A rectangle is cut diagonally to form 2 e

qual triangles. How do the height and base length of the two triangles compare to the height and base length of the original piece of paper?

1 answer:

Zanzabum3 years ago

8 0

Answer:

The two dimensions are the same in the triangles as they were in the rectangle.

Step-by-step explanation:

Height of a triangle is the perpendicular distance from the one vertex of the triangle to the one side of the triangle( that side is called base of the triangle).

Let the length of the given rectangle is l unit and width of the rectangle is b unit,

When the rectangle is cut along its diagonal,

Then we found two right triangles,

In which both having dimensions,

height = b and base = l,

Thus, the dimension of the triangles are same as the rectangles.

⇒ First option is correct.

Read more on Brainly.com - brainly.com/question/12150549#readmore

You might be interested in

How do I Determine if ratios are equivalent

Veseljchak [2.6K]

Step-by-step explanation:

multiply the ratio by the second ratios second number

4 0

3 years ago

Pls answer i will fail:-}

True [87]

Answer:

Part 1:

You can solve the pair of equation graphically by writing the y-intercepts which are (0,-1) and (0,-5) Then follow the slopes of each line till the intercept. The slopes are up 2 over 1 and up 4 over 1 respectivily

Part 2:

When you graph these you get the answer of (2,3)

x is 2 and y is 3

Step-by-step explanation:

Hope it helps :)

pls mark brainliest :P

*insert British accent* "Don't just let 'em take it!"

3 0

2 years ago

Read 2 more answers

Simplify to get the right answer

Musya8 [376]

The correct answer would be: 18+18i

5 0

4 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

4 years ago

Evaluate the response...

Kruka [31]

Answer:

um 18 lol

Step-by-step explanation:

Please mark brainliest

4 0

3 years ago

Read 2 more answers