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3 years ago

Which of these pairs of events are independent?

1 answer:

3 0

B.You reach into a basket and draw a name for a prize, return the name into the basket, and then draw a second time for another prize.

The results of the prizes in the first basket are not related at all to what happens in the second basket. So these two events are independent.

Hope this helps :)

You might be interested in

How do you find the surface area of this figure?

atroni [7]

Answer:

SA=158cm²

Step-by-step explanation:

To find surface area, you find the area for all the surfaces and add them together. Let's start with the red rectangular prism first.

The top has measurements of 5 and 2. The area of that is 10.

The bottom is the same so it's area is also 10 but we don't want to add it since it's not on the surface.

The side facing us has measurements of 4 and 2, the area is 8.

The side opposite to this is the same so it's area is also 8

Now the right face of this shape is 4 and 4 so 16.

The left face will have the same area. It'll also be 16.

The surface area of the red shape is 10+8+8+16+16= 58

Now the orange shape.

The side facing us is 5 and 3 so area=15

The side opposite to this will have the same area.= 15

The right face is also 5 and 3 so 5x3=15

The opposite face to this will also have an area of 15

Now the top face, it's 5 and 5 = 25. We have to subtract the area covered by the red shape. So 25 - 10 = 15

The bottom face's area will be 25.

The surface area of this shape will be : 15+15+15+15+25=100

Now add the areas of both shapes

58+100 = 158

SA=158cm²

Hope this helps :)

4 0

3 years ago

Kevin claims he can draw a trapezoid whit three right angles. is this posible (explain)

Zielflug [23.3K]

No because if u count the corners there are 4 right angles

3 0

3 years ago

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .