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ruslelena [56]
3 years ago
10

Are the ratios 2.5:3.5 and 5.7 proportional ?

Mathematics
1 answer:
romanna [79]3 years ago
3 0
Yes, they are proportional, as when you double 2.5, you get 5, and when you double 3.5, you get 7
Hope this helps :) 
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Is there a proportional relationship between x and y? Explain.
Vlad1618 [11]
I don’t think so but divide the top one by the bottom one for every one of them and if it is all the same answer they are proportional x/y
4 0
3 years ago
-2x+3=15? What is the answer
const2013 [10]
-2x+3=15

Subtract 3 from both sides:

-2x=12

Divide -2 from both sides:

X=-16 is your answer
5 0
3 years ago
the bottom of a ladder must be placed 3 feet from a wall the laser is 10 feet long how far above the ground does the ladder touc
kaheart [24]
It would be 7 feet above ground
6 0
3 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
Philip's family traveled 3/10 of the distance to his grandmother's house on saturday. They traveled 4/7 of the remaining distanc
Nataly [62]
Okay do 4*10 and 7*10 (you get 40/70) and then do 3*7 and 10*7 (so now you've got 21/70) then add 21/70+40/70 and that's the total distance
6 0
3 years ago
Read 2 more answers
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