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3 years ago

15

A bag contains 4 yellow marbles and 6 green marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then

a second marble is drawn. What is the probability that the first marble is green and the second one is yellow?
1/15

4/15

1/6

3/2

1 answer:

Pie3 years ago

7 0

$\frac{6}{10} \times \frac{4}{9} \\ = \frac{24}{90} \\ = \frac{4}{15}$

Answer is 2nd option, 4/15

Hope this helps. - M

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Find the measure of angle a.

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a = 127°

Step-by-step explanation:

The 3 angles lie on a straight line and have a sum of 180°, that is

15° + a + 38° = 180°

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a = 127°

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4 years ago

Find the rectangular coordinates of the point with the polar coordinates (3, 3 divided by 2 pi ). (1 point)

kifflom [539]

Answer:

The correct choice is (0,-3)

Step-by-step explanation:

To convert from polar coordinates to rectangular coordinates, we use the parametric equations:

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The point given to us in polar coordinates is $(3,\frac{3}{2}\pi)$ .

where $r=3,\theta=\frac{3\pi}{2}$

$x=3\cos (\frac{3\pi}{2})$ and $y=3\sin( \frac{3\pi}{2})$

$x=3(0)$ and $y=3(-1)$

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3 years ago

The moon is a sphere with radius of 959 km. Determine an equation for the ellipse if the distance of the satellite from the surf

Sergeeva-Olga [200]

Answer:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

Step-by-step explanation:

An ellipse is the locus of a point such that its distances from two fixed points, called foci, have a sum that is equal to a positive constant.

The equation of an ellipse with a center at the origin and the x axis as the minor axis is given by:

$\frac{x^2}{b^2}+\frac{y^2}{a^2} =1 \\\\where\ a>b$

Since the distance of the satellite from the surface of the moon varies from 357 km to 710 km, hence:

b = 357 km + 959 km = 1316 km

a = 710 km + 959 km = 1669 km

Therefore the equation of the ellipse is:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

5 0

3 years ago

PLEASE PLEASE ANSWER!

muminat

The area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = $\frac{1}{2}\times \pi r^2$

$$=\frac{1}{2}\times \pi\times 4^2$

$$=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $$8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

$$=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC = $8\sqrt{3}$ cm²

Area of the shaded region

= Area of the semi-circle – Area of the triangle ABC

= $$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

= $$8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

3 0

3 years ago