All categories

3 years ago

A small pouch contains 8 black marbles,5 white marbles, and 12 red marbles. Find the probability of NOT picking a red marble

Mathematics

1 answer:

Elis [28]3 years ago

5 0

Answer:

10/36

Step-by-step explanation:

You might be interested in

Which graph shows a proportional relationship ?

morpeh [17]

It's graph A because it going threw the and has a pattern.

4 0

3 years ago

A square has a diagonal length of 10 ft. Draw a diagram and determine the perimeter of the square.

denpristay [2]

The perimeter of the square is: $20\sqrt2$ ft

Step-by-step explanation:

The picture is attached with the answer.

the diagonal of a square divides it into two right angled-triangles.

We can use one of the triangle to find the length of side as the diagonal will be the hypotenuse of the triangle

Let a be the side of the square, then according to Pythagoras theorem

$d^2=a^2+a^2\\(10)^2 = 2a^2\\2a^2=100$

Dividing both sides by 2

$\frac{2a^2}{2}=\frac{100}{2}\\a^2=50$

Taking Square root on both sides

$\sqrt{a^2}=\sqrt{50}\\a=\sqrt{25*2}\\a=\sqrt{5^2 * 2}\\a=5\sqrt{2}\ ft$

Now,

$P = 4a\\P = 4 * 5\sqrt{2}\\=20\sqrt{2}\ ft$

Hence,

The perimeter of the square is: $20\sqrt2$ ft

Keywords: Perimeter, Pythagoras Theorem

Learn more about Square at:

#learnwithBrainly

5 0

3 years ago

What is the general equation of a cosine function with an amplitude of 3, a period of 4pi, and a horizontal shift of -pi?

HACTEHA [7]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Reflect (3,9) across the y-axis.

Andru [333]

Answer:

(-3,-9)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.

Oduvanchick [21]

Answer:

$11402.66 m^{2}$

Step-by-step explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by $\frac {0.5\pi d^{2}}{4}$

Total area of semi circle will be $2\times\frac {0.5\pi d^{2}}{4}$

Substituting 74 m for d and $\pi$ as 3.14 we obtain

Total area semi-circle= $2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}$

Area of rectangle is given by the product of length and width

Rectangular area= $96 m*74 m=7104 m^{2}$

Total area of rectangular and semi-circles will be

$4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}$

Therefore, area of training field is $11402.66 m^{2}$

6 0

3 years ago