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katrin [286]
2 years ago
11

Look at the shaded shapes drawn on a square grid.

Mathematics
1 answer:
Triss [41]2 years ago
4 0

Answer:

No.  Steven is not correct.

Step-by-step explanation:

The area of shape A is 15 (either multiply 5 x 3 or just count squares).  The area of shape B is 9.

15 is not double 9.

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You are very welcome
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0.12 word form decimal
7nadin3 [17]

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0.12=zero point one two

Step-by-step explanation:

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Yesterday, 12 kids played Among Us and 8 kids did
marysya [2.9K]

Answer:

60%

Step-by-step explanation:

12 kids played. 8 didn't play. That means there is a total of 20 students.

The fraction would be 12/20.

To make it into a percent, divide 100 by the denominator (20) and multiply it by the numerator (12)

100 divided by 20 is 5

12 x 5 = 60

5 0
2 years ago
Which integer is closest to 0 on the number line?
Elenna [48]
The answer is C, 1.
In a number line, the larger number it gets, the more of the right side they're. The smaller the number, the more left side they get.
So, on a number line, 1 is just at the right side of 0, all the other options are at least one more place away from 0.
7 0
3 years ago
Read 2 more answers
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
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