The sum of 7 and the cube of a number as an equation

1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:

A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?

You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked

2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.

b - 1,000,000/2500 = 400
400 packages are defective each year

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3 years ago

The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a fo

qwelly [4]

1. The distance between the perihelion and the aphelion is 116 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Step-by-step explanation:

Let us revise the equation of the ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where the major axis is parallel to the x-axis

The length of the major axis is 2a
The coordinates of the vertices are (± a , 0)
The coordinates of the foci are (± c , 0) , where c² = a² - b²

∵ The Sun is located at a focus of the ellipse

∴ The sun located ate c

∵ The perihelion is the point in a planet’s orbit that is closest to the

Sun ( it is the endpoint of the major axis that is closest to the Sun )

∴ The perihelion is located at the vertex (a , 0)

∵ The closest Mercury comes to the Sun is about 46 million miles

∴ The distance between a and c is 46 million miles

∵ The aphelion is the point in the planet’s orbit that is furthest from

the Sun ( it is the endpoint of the major axis that is furthest from

the Sun )

∴ The aphelion is located at the vertex (-a , 0)

∵ The farthest Mercury travels from the Sun is about 70 million miles

∴ The distance from -a to c is 70 million miles

∴ The distance between the perihelion and the aphelion =

70 + 46 = 116 million miles

1. The distance between the perihelion and the aphelion is 116 million miles

∵ The distance between the perihelion and the aphelion is the

length of the major axis of the ellipse

∵ The length of the major axis is 2 a

∴ 2 a = 116

- Divide both sides by 2

∴ a = 58

∴ The distance from the center of Mercury’s elliptical orbit to the

closest end point to the sun is 58 million miles

∵ The distance between the sun and the closest endpoint is

46 million miles

∴ The distance from the center of Mercury’s elliptical orbit and

the Sun = 58 - 46 = 12 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

∵ The major axis runs horizontally

∴ The equation is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

∵ a = 58

∵ c is the distance from the center to the focus of the ellipse

∴ c = 12

∵ c² = a² - b²

∴ (12)² = (58)² - b²

- Add b² to both sides

∴ (12)² + b² = (58)²

- Subtract (12)² from both sides

∴ b² = (58)² - (12)² = 3220

- Substitute these values in the equation

∴ $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

The eccentricity (e) of an ellipse is the ratio of the distance from the

center to the foci (c) and the distance from the center to the

vertices (a) ⇒ $e=\frac{c}{a}$

∵ c = 12

∵ a = 58

∴ $e=\frac{12}{58}$ = 0.207

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

If the eccentricity is zero, it is not squashed at all and so remains a circle.

If it is 1, it is completely squashed and looks like a line

∵ The eccentricity of the ellipse is 0.207

∵ This number is closed to zero than 1

∴ The shape of the ellipse is near to the shape of the circle

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Learn more:

You can learn more about conics section in brainly.com/question/4054269

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3 years ago