All categories

4 years ago

Complete the conditional statement if -2 a>6 then A. A>3 B. A <3 C. A <-3

1 answer:

8 0

The correct answer is C.) A < -3
 
I would have explained why, but I have to go.

You might be interested in

Yesterday, the snow was 2 feet deep in front of Archie’s house. Today, the snow depth dropped to 1.6 feet because the day is so

IgorLugansk [536]

Answer: The depth of the snow has changed .4 inches. This is because it has melted in the warm weather. You can find the difference, by subtracting 1.6 from 2.

Step-by-step explanation:

2 - 1.6= .4 .4 would be your answer

Hope it helps!!!!

4 0

3 years ago

Find the volume of a cylinder with radius 5.7cm and height 10 cm, correct to 1 decimal place

Veseljchak [2.6K]

$\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5.7\\ h=10 \end{cases}\implies \begin{array}{llll} V=\pi (5.7)^2(10)\implies V=324.9\pi \\\\\\ V\approx 1020.7 \end{array}$

8 0

2 years ago

Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first,

Kryger [21]

Answer:

$E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}$

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value $x \in (0,1)$ we want this:

$max(U_1, ....,U_n) \leq x$

And we can express this like that:

$u_i \leq x$ for each possible i

We assume that the random variable $u_i$ are independent and $P)U_i \leq x) =x$ from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

$P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n$

And then cumulative distribution would be expressed like this:

$0, x \leq 0$

$x^n, x \in (0,1)$

$1, x \geq 1$

For each value $x\in (0,1)$ we can find the dendity function like this:

$f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}$

So then we have the pdf defined, and given by:

$f_X (x) = n x^{n-1} , x \in (0,1)$ and 0 for other case

And now we can find the expected value for the random variable X like this:

$E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}$

$E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}$

6 0

3 years ago

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this

Firlakuza [10]

The amount of substance left of a radioactive element of half life, $t_{\frac{1}{2}}$ after a time, t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}$

Given that potassium-40 has a half life of approximately 1.25 billion years.

The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:

$0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5$

Therefore, the maximum age of a fossil that we could date using 40k is 12.5 billion years.

3 0

3 years ago

. Express the following as a percent: 5 out of 30. Round to the nearest whole number

balandron [24]

5,10,15,20,25,30 that is 1,2,3,4,5,6 6 of 100 is 18

3 0

3 years ago

Read 2 more answers