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3 years ago

Someone help me with 27

Mathematics

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

The answer is $(x^3-8)(x^3+2)$

Step-by-step explanation:

Find two factors that equal -6 when added, and -16 when multiplied. Those are your outer numbers. The exponent of the first term is a 6, so you know when you multiply two numbers with exponents, you add them. 6/2=3.

I'm really sorry for the bad explanation, I was in a hurry. Sorry. But I know the answer is correct.

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What is 152,909 rounded to the nearest hundred thousand

marta [7]

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David was in need for some money so he decided to clean out the water fountain at the mall he found 20 nickels and quarters. The

Shkiper50 [21]

Answer:

David find 8 quarters.

Step-by-step explanation:

Given:

David was in need for some money so he decided to clean out the water fountain at the mall he found 20 nickels and quarters.

The collection of nickels and quarters totaled to 2.6.

Now, to find the quarters David find.

Let the number of nickels be $x$

And the number of quarters be $y$

Thus, the number of nickels and quarters he found:

$x+y=20.$

⇒ $x=20-y$ .........( 1 )

The totaled of nickels and quarters:

$0.05x+0.25y=2.6$

(As 1 nickel=$0.05 and 1 quarter=$0.25)

Now, putting the equation ( 1 ) in the place of $x$ we get:

$0.05(20-y)+0.25y=2.6$

⇒ $1-0.05y+0.25y=2.6$

⇒ $1+0.20y=2.6$

Subtracting both sides by 1 we get:

⇒ $0.20y=1.6$

Dividing both sides by 0.20 we get:

⇒ $y = 8.$

So, the quarters = 8.

Therefore, David find 8 quarters.

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3 years ago

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Masteriza [31]

The answer is 43 hope this helps:)

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Let XX be a random variable that is equal to the number of heads in two flips of a fair coin. What is \text E[X^2]E[X 2 ]

Anna [14]

Answer:

Step-by-step explanation:

From the given information, it is likely that the random variable(X) have the values below:

Let head be H

Let tail be T

So;

X(HH) = 2;

X(HT) = 1;

X(TH) = 1;

X(TT) = 0

The distribution can now be computed as:

$p(X= TT) = \dfrac{1}{4}$

$p(X=TH) = \dfrac{1}{4}$

$p(X=HT) = \dfrac{1}{4}$

$p(X=HH)= \dfrac{1}{4}$

Now, the expected value that is equivalent to the number of heads when the coin is flipped twice is:

$E(X) = p(TT)*X(TT)+p(TH)*X(TH)+p(HT)*X(HT)+p(HH)*X(HH)$

$E(X) = \dfrac{1}{4}\times 0 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 2$

$E(X) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{1}{2}$

$E(X) =\dfrac{1+1+2}{4}$

$E(X) =\dfrac{4}{4}$

E(X) = 1

$E(X^2) = p(TT)*X(TT)^2+p(TH)*X(TH)^2+p(HT)*X(HT)^2+p(HH)*X(HH)^2$

$E(X^2) = \dfrac{1}{4}\times 0^2+ \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 2^2$

$E(X^2) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{4}{4}$

$E(X^2) =\dfrac{1+1+4}{4}$

$E(X^2) =\dfrac{6}{4}$

$E(X^2) =1.5$

Finally; To compute E²[X]

E²[X] = E[X]²

E²[X] = 1²

E²[X] = 1

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3 years ago