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2 years ago

10

A contestant on a game show must choose 2 of 16 boxes. Once a box is chosen, it is removed from the game. Of the 16 boxes, 3 con

tain money, 5 contain traveling prizes, and 8 are empty. What is the probability that both boxes are empty?

1 answer:

Inessa05 [86]2 years ago

7 0

Answer:

7 / 30

Step-by-step explanation:

This is a selection without replacement probability :

Total boxes = 16

With money = 3

With traveling prices = 5

Empty = 8

P(choosing 2 empty boxes) :

P(empty) * P(empty)

P(empty) = Number of empty boxes / total number of boxes

First pick = empty

P(empty) = 8 / 16 = 1/2

Without replacement :

Second pick = empty

P(empty) = 7 / 15

P(choosing 2 empty boxes) = 1/2 * 7/15 = 7 /30

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Identify the equivalent expression for each of the expressions below (m1/3 m1/5)^0

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Answer:

1

Step-by-step explanation:

everything rise 0 power is 1.

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3 years ago

50% of all the cakes Lucy bakes that week were party cakes: one fifth were fruit cakes and the remainder were sponge cakes. What

dusya [7]

50% is equal to 5/10 so 5/10 were party cakes 1/5x2=2/10 so 2/10 were fruit cakes 5/10+2/10=7/10 so 10/10 which is 100% of all cakes 10/10-7/10=3/10 so 3/10 of the cakes were sponge cakes.

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3 years ago

Suppose you have 5 riders and 5 horses, and you want to pair them off so that every rider is assigned one horse (and no horse is

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There are 120 ways in which 5 riders and 5 horses can be arranged.

We have,

5 riders and 5 horses,

Now,

We know that,

Now,

Using the arrangement formula of Permutation,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$ ,

So,

For n = 5,

And,

r = 5

As we have,

n = r,

So,

Now,

Using the above-mentioned formula of arrangement,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$ ,

Now,

Substituting values,

We get,

$^5N_5 = \frac{5!}{(5-5)!}$

We get,

The total number of ways of arrangement = 5! = 5 × 4 × 3 × 2 × 1 = 120,

So,

There are 120 ways to arrange horses for riders.

Hence we can say that there are 120 ways in which 5 riders and 5 horses can be arranged.

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2 years ago

Perform the indicated operation and express the result as a simplified complex number. I need help with 35

skelet666 [1.2K]

Given:-

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To find:-

The simplified form.

At first we take conjucate and multiply below and above.

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$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,

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So the required solution is,

$\frac{2}{3}+i\frac{11}{3}$

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1 year ago

Evaluate 2x for each value of x. Write your answers in simplest form. Show your work!

Liono4ka [1.6K]

1. 1/2

2.2/3

3. 1

4. 1/3

5. 2/7

6. 1/4

7. 4/3

8. 3/2

<h3>How to evaluate the value</h3>

To find 2x of the vale of x, we have to multiply the value of 'x' by 2

1. x = 1/4

$2x = 2 * \frac{1}{4}$ ⇒ $\frac{2}{4}$ ⇒ $\frac{1}{2}$

2. x = 1/3

$2x = 2 * \frac{1}{3}$ ⇒ $\frac{2}{3}$

3. x = 1/2

$2x = 2* \frac{1}{2}$ ⇒ $\frac{2}{2}$ ⇒ $1$

4. x= 1/6

$2x = 2* \frac{1}{6}$ ⇒ $\frac{2}{6}$ ⇒ $\frac{1}{3}$

5. x = 1/ 7

$2x = 2 * \frac{1}{7}$ ⇒ $\frac{2}{7}$

6. x = 1/8

$2x = 2 * \frac{1}{8}$ ⇒ $\frac{2}{8}$ ⇒ $\frac{1}{4}$

7. x = 2/3

$2x = 2 * \frac{2}{3}$ ⇒ $\frac{4}{3}$

8. x = 3/4

$2x = 2 * \frac{3}{4}$ ⇒ $\frac{6}{4}$ ⇒ $\frac{3}{2}$

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2 years ago