Ask question

Ask question

All categories

1 year ago

8

Each of ten tickets is marked with a different number from 1 to 10 and put in a box. If you draw a ticket from the box, what is

the probability than
you will draw a number greater than 3?

1 answer:

Ahat [919]1 year ago

6 0

Answer:

the probability is 7

Step-by-step explanation:

if you think about it this way you have 10 tickets:

1 2 3 4 5 6 7 8 9 10

they say that if you pull anything greater than 3 so anything greater than 3 would be higher. 4, 5, 6, 7, 8, 9, 10 if you count from 4- 10 the answer is 7. I don't know if that helped at all but hopefully it did.

You might be interested in

Which pairs of quadrilaterals can be shown to be congruent using rigid motions?

Misha Larkins [42]

Answer:

1 and 2 are not congruent; 1 and 3 are congruent; 1 and 4 are congruent; 2 and 3 are not congruent; 2 and 4 are not congruent; 3 and 4 are congruent.

Step-by-step explanation:

From the diagram, we can see that the angle measures and side measures of figures 1, 3 and 4 are the same. This means that these three figures are congruent. Figure 2, however, is not congruent to any of the other 3.

5 0

3 years ago

Read 2 more answers

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

2 years ago

What is the solution to the equation -2(1+5a)=-6(a+3) ?

Lilit [14]

The solution is a=4

-2(1+5a)=-6(a+3)

-2-10a=-6a-18

+18 +18

16-10a=-6a

+10a +10a

16=4a

divide by 4

4=a

7 0

2 years ago

How do solve p+5=-2 with steps

Basile [38]

P+5= -2

p= -7 ( I subtracted 5 to both sides of the equation)

4 0

3 years ago

| -18 + 1| absolute value and helppplpllll

muminat

Answer:

17

Step-by-step explanation:

| -18 + 1|

-18 + 1 = -17

So,

| -17| = 17

This is because if | -a|, then | -a| = a

6 0

2 years ago

Read 2 more answers