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3 years ago

How to use replacement in probability

Mathematics

1 answer:

dolphi86 [110]3 years ago

7 0

Step-by-step explanation:

Replacement means the probability of each trial is the same (the trials are independent).

For example, let's say you have a standard deck of 52 cards, and you want to find the probability of drawing a queen card twice.

There are 4 queens, so the probability on the first draw is 4/52.

You then replace the queen, so there are still 4 queens in the deck. So the probability on the second draw is still 4/52.

The total probability is the product: 4/52 × 4/52 = 1/169

If you didn't replace the queen after the first draw, there would have been 3 queens in the deck, and the probability on the second draw would have been 3/52.

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54 over 100 in simplest form

alex41 [277]

It would be 27/50
54/2 = 27
100/2 = 50

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3 years ago

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FACTOR....<br>x^2 + 10× - 2400 = 0

Vikki [24]

Answer:

x= -5 + 5 $\sqrt{97}$ , x= -5 - 5 $\sqrt{97}$

Step-by-step explanation:

Since this quadratic is set to zero, we can use the quadratic formula to solve this.

x^2 + 10x - 2400 = 0

Quadractic formula = x= -b +- $\sqrt{b^2 - 4ac}$ /2a

For this equation:

a= 1, b=10, c=-2400

Plug these numbers into the equation and solve.

x= -10 +- $\sqrt{10^2 - 4(1)(-2400}$ )/2(1)

x= -10 +- $\sqrt{100 + 9,600}$ /2

x= -10 +- $\sqrt{9,700}$ /2

x= -10 +- $\sqrt{2^2 * 5^2 * 97}$ /2

x= -10 +- 5 * 2 $\sqrt{97}$ /2

x= -10 +- 10 $\sqrt{97}$ / 2

Divide by 2.

x= -5 +- 5 $\sqrt{97}$

Answer:

x= -5 + 5 $\sqrt{97}$ or x= -5 - 5 $\sqrt{97}$

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3 years ago

Each day Mrs. Yoder assigns 15 addition and 12 multiplication problems as homework. What is the total number of addition and mul

schepotkina [342]

Answer:

27 each day

Step-by-step explanation:

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3 years ago

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Please explain your answer

Irina18 [472]

$\text{Distance = } \sqrt{(5 + 8)^2+ (-3 + 3)^2 }= \sqrt{13^2} = 13$

Answer: 13 units

6 0

3 years ago

What are all values of x for which the series shown converges?

adoni [48]

Answer:

One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:

Iaₙ₊₁I < IaₙI

This means that the absolute value of the terms must decrease as n increases.

Then we must have:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}}$

We can write this as:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}} = \frac{(x -2)^n}{(n + 1)*3^n} * \frac{(x - 2)}{3}$

If we assume that n is a really big number, then:

n + 1 ≈ 1

And we can write:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2)^n}{(n)*3^n} * \frac{(x - 2)}{3}$

Then we have the inequality

$1 > (x - 2)/3$

And remember that this must be in absolute value, then we will have that:

-1 < (x - 2)/3 < 1

-3 < x - 2 < 3

-3 + 2 < x < 3 + 2

-1 < x < 5

The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.

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3 years ago