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3 years ago

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There are 5 slots, each containing the letters W, R, L, D or O. One letter is picked at random from each slot. What are the odds

that the letters stored in these slots read the word WORLD?

1 answer:

NARA [144]3 years ago

5 0

Answer:

1/120

Step-by-step explanation:

For the first letter, you have a 1/5 chance of getting w

On the second you have a 1/4 chance to get the r

Then 1/3 and 1/2

Next you just multiply the bottom numbers

That gives you how many diffrent outcomes there can be. Put that over 1 and you have your answer.

Hope this helps <3

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Answer:

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2 years ago

What is the probability of drawing the compliment of a king or a

inna [77]

Answer:

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Let 'S' be the sample space associated with the drawing of a card

n (S) = 52C₁ = 52

Let E₁ be the event of the card drawn being a king

$n( E_{1} ) = 4 _{C_{1} } = 4$

Let E₂ be the event of the card drawn being a queen

$n( E_{2} ) = 4 _{C_{1} } = 4$

But E₁ and E₂ are mutually exclusive events

since E₁ U E₂ is the event of drawing a king or a queen

<u><em>step(ii):-</em></u>

The probability of drawing of a king or a queen from a standard deck of playing cards

P( E₁ U E₂ ) = P(E₁) +P(E₂)

$= \frac{4}{52} + \frac{4}{52}$

P( E₁ U E₂ ) = $\frac{8}{52}$

<u><em>step(iii):-</em></u>

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards

$P(E_{1}UE_{2}) ^{-} = 1- P(E_{1} U E_{2} )$

$P(E_{1}UE_{2}) ^{-} = 1- \frac{8}{52}$

$P(E_{1}UE_{2}) ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846$

<u><em>Conclusion</em></u>:-

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

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3 years ago

Can someone help me with this please

11Alexandr11 [23.1K]

Answer:

I have no clue

Step-by-step explanation:

1. Search online

2. Never count on me again!

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3 years ago

Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.

Levart [38]

A factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

<h3>What are the properties of roots of a polynomial?</h3>

The maximum number of roots of a polynomial of degree $n$ is $n$ .
For a polynomial with real coefficients, the roots can be real or complex.
The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if $a+ib$ is a root, then $a-ib$ is also a root.

If the roots of the polynomial $p(x)=ax^4+bx^3+cx^2+dx+e$ are $r_1,r_2,r_3,r_4$ , then it can be factorized as $p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ .

Here, we are to find a factorization of $p(x)=x^4+2x^3+7x^2-6x+44$ . Also, given that $-2+i\sqrt{7}$ and $1-i\sqrt{3}$ are roots of the polynomial.

Since $p(x)=x^4+2x^3+7x^2-6x+44$ is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, $-2-i\sqrt{7}$ and $1+i\sqrt{3}$ are also roots of the given polynomial.

Thus, all the four roots of the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ , are: $r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}$ .

So, the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ can be factorized as follows:

$\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)$

Therefore, a factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

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1 year ago

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A girl was 36 inches tall at age 4. At at 6, she was 41 inches tall.

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48.5 inches is the answer

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