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erik [133]
3 years ago
7

A jar contained 8 balls: 6 red, 1 green and 1 blue. We randomly take a ball out of the jar. Once a ball is selected we do not pu

t it back. We keep selecting balls until for the first time we get a blue one. We count the number of times we had to draw a ball, (including the last drawing) until we actually got the blue ball. How many elements are in the sample space for this experiment
Mathematics
1 answer:
Oxana [17]3 years ago
8 0

Answer:

35

Step-by-step explanation:

Given that :

Number of balls = 8

Red (R) = 6 ; 1 green(G) ; 1 BLUE (B)

Possibilities unt a blue ball is picked :

B, RB, GB, RRB, RGB, GRB, G

Draw 1 = B = 1

Draw 2 = B = 2C1 = 2

Draw 3 = B = 3C2 = 3

Draw 4 = B = 3C3 + 3C1 = 1 + 3 = 4

Draw 5 = B = 4C4 + 4C1 = 1 + 4 = 5

Draw 6 = B = 5C5 + 5C1 = 1 + 5 = 6

Draw 7 = B = 6C6 + 6C1 = 1 + 6 = 7

Draw 8 = B = 6C6 + 6C1 = 1 + 6 = 7

Taking the sum:

(1 + 2 + 3 + 4 + 5 + 6 + 7 + 7) = 35

There are 35 elements in the sample space

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Answer:

Volume: 52 Units Squared

Surface Area: 94 units.

Step-by-step explanation:

The volume is relatively simple to find. Just subtract the original volume by the 2x2x2 cube's volume. The original volume is 60. The cube's volume is 2x2x2 which is 8. 60-8=52.

The surface area is harder to find. Try to envision the corner of the rectangle being cut out. We see that each side of the cube has a surface area of 2x2 which is 4. In the picture, we see that three sides of the rectangle has been partially removed. But since each side of the cube has an equal surface area, it is safe to minus 3 of the sides that has been partially removed by 3. However, since that it is the corner, the "dent" that the cube made in the rectangle also needed to be counted. As we said, each of the sides of a cube has a surface area of 4, so since that the dent has 3 sides, we see that the surface area of the dent is 4x3 which is 12. Now we need to count the unaffected sides of the rectangle. There are three of them. Just multiply the edges to find the surface area of each side. Add all of the values up: 11+16+12+8+15+12+20=94 units.

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How can the difference 13-(-24)be rewritten as a sum with the correct solution? Explain?
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Read 2 more answers
Factorise 2x^3+9x^2+10x+3​
Advocard [28]

In order to factor a polynomial p(x), we have to find all its roots x_1,\ x_2,\ldots x_n so that we can rewrite the polynomial as

p(x)=(x-x_1)(x-x_2)\ldots(x-x_n)

This is exactly the same idea we apply when we factor numbers: we look for all the primes that divide the number, and then we write

n = p_1^{e_1}\cdot p_2^{e_2}\ldots p_n^{e_n}

When talking about polynomials, the idea of prime numbers is represented by irreducible polynomials, i.e. polynomials with no roots.

So, we have to find a root of our polynomial. Using the rational root theorem, we can check that x=-1 is a solution:

p(-1)=2(-1)^3+9(-1)^2+10(-1)+3 = -2+9-10+3=0

So, our polynomial is divisible by (x+1). The long division yields

\dfrac{2x^3+9x^2+10x+3​}{x+1} = 2x^2+7x+3

Which is the same as

2x^3+9x^2+10x+3=(x+1)(2x^2+7x+3)

We can complete the factorization by breaking the quadratic equation: using the standard quadratic formula we can find the solutions

2x^2+7x+3=0 \iff x=-\dfrac{1}{3}\text{ or }x=-3

Which implies

2x^2+7x+3=\left(x+\dfrac{1}{2}\right)(x+3)

And finally

2x^3+9x^2+10x+3=(x+1)\left(x+\dfrac{1}{2}\right)(x+3)

8 0
3 years ago
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