George deposited $900 in an account that earns 5% annual compounded annually. How much

Graph quadrilateral ABCD with vertices A(2,-3), B(3,5), C(6,6) and D(5,-3) and its image after the translation (x,y)→(x−5,y−2)

maria [59]

The graph can be drawn by the placing the points and after translation points becomes A'(-3, -5), B'(-2, 3), C'(1, 4), and D'(0, -5)

<h3>What is quadrilateral?</h3>

It is defined as the four-sided polygon in geometry having four edges and four corners.

We have:

Quadrilateral ABCD with vertices A(2,-3), B(3,5), C(6,6) and D(5,-3)

We can graph it on the coordinate plane as shown in the picture.

After applying the translation (x, y)→(x−5, y−2), the points becomes:

A'(-3, -5)

B'(-2, 3)

C'(1, 4)

D'(0, -5)

Similarly, we can graph it on the coordinate plane (refer attached picture)

Thus, the graph can be drawn by the placing the points and after translation points becomes A'(-3, -5), B'(-2, 3), C'(1, 4), and D'(0, -5)

Learn more about the quadrilateral here:

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6 0

2 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

Y=\sqrt(-x^(2)+2x) . tìm giá trị lớn nhất của hàm số

Jlenok [28]

Answer:

UM hello how are you im njot really gonna answer anywaysss byee

3 0

2 years ago

The difference between a number and 7 is greater than -23

Aleks04 [339]

-23 is a negative number so that means 7 is a greater number. any number with a minus sign like this - means it’s a negative number, so any number that you have to compare it to and if that number has nothing before or after it, then that would be the greater number

5 0

2 years ago

Ricardo is constructing the inscribed circle for △ABC . He has already used his compass and straightedge to complete the part of

Umnica [9.8K]

<span>Construct the perpendicular to BC¯¯¯¯¯ through point R.</span>

7 0

3 years ago

Read 2 more answers