Round this number to the nearest 10,000. 64,195 A. 70,000 B. 50,000 C. 60,000

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

We can conclude that the change to the face 3 affects the value of P(Odd Number)

7 0

2 years ago

Two column proof:

gulaghasi [49]

Answer:

Given : BRDG is a kite that is inscribed in a circle,

With BR = RD and BG = DG

To prove : RG is a diameter

Proof:

Since, RG is the major diagonal of the kite BRDG,

By the property of kite,

∠ RBG = ∠ RDG

Also, BRDG is a cyclic quadrilateral,

Therefore, By the property of cyclic quadrilateral,

∠ RBG + ∠ RDG = 180°

⇒ ∠ RBG + ∠ RBG = 180°

⇒ 2∠ RBG = 180°

⇒ ∠ RBG = 90°

⇒ ∠ RDG = 90°

Since, Angle subtended by a diameter or semicircle on any point of circle is right angle.

⇒ RG is the diameter of the circle.

Hence, proved.

8 0

3 years ago

You want to buy three books that are on sale at 20% off. The original prices of the books are $2.50, $4.95, and $6.00. How much

Anon25 [30]

First add the prices of the original books.

2.5+4.95+6=11.45

Find 20% of 11.45

=2.29

Therefore I saved $2.29

8 0

3 years ago

Thank you to whoever helps!

shepuryov [24]

Answer:

M
90°

Step-by-step explanation:

The segment joining an original point with its rotated image forms a chord of the circle of rotation containing those two points. The center of the circle is the center of rotation.

This means you can find the center of rotation by considering the perpendicular bisectors of the segments joining points with their images. Here, the only proposed center that is anywhere near the perpendicular bisector of DE is point M.

__

Segment AD is perpendicular to corresponding segment FE, so the angle of rotation is 90°. (We don't know which way (CW or CCW) unless we make an assumption about which is the original figure.)

4 0

3 years ago

There are 2 sets of balls numbered l through 19 placed in a bowl. If 2 balls are randomly chosen without replacement, find the p

Sergio039 [100]

Answer:

0.027027

Step-by-step explanation:

2 x 19= 38

38C2=703

19/703=0.027027

You

8 0

2 years ago