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Mashutka [201]
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?
Mathematics
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.

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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
Item 7
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
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