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3 years ago

Find two numbers whose sum is -39 and whose product is 10

1 answer:

3 0

A) x + y = -39
B) x * y = 10

From A) we see x = -y -39 then we substitute this into B
B) (-y -39) * y = 10
-y^2 -39y = 10
-y^2 -39y -10 = 0

y = -38.742
x = -.258

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In the coordinate plane, the point A (-2,-1) is translated to the point A (-4,0). Under the same translation, the points B (-4,2

Zolol [24]

Answer:

The coordinates for B' would be (-6,1) and the coordinates for C would be (-8,-6)

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3 years ago

If x=10, write an expression in terms of x for the number 5,364

LenKa [72]

Answer:

(5,354 + x)

536.4*x

Step-by-step explanation:

We know that x = 10.

Now we want to write an expression (in terms of x) for the number 5,364.

This could be really trivial, remember that x = 10.

Then: (x - 10) = 0

And if we add zero to a number, the result is the same number, then if we add this to 5,364 the number does not change.

5,364 = 5,364 + (x - 10) = 5,364 + x - 10

5,364 = 5,354 + x

So (5,354 + x) is a expression for the number 5,364 in terms of x.

Of course, this is a really simple example, we could do a more complex case if we know that:

x/10 = 1

And the product between any real number and 1 is the same number.

Then:

(5,364)*(x/10) = 5,364

(5,364/10)*x = 5,364

536.4*x = 5,364

So we just found another expression for the number 5,364 in terms of x.

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3 years ago

Each letter of the word "supercalifragilisticexpialidocious" is placed into a bag and drawn at 3 times, replacing the letter aft

Makovka662 [10]

Answer:

P(X ≥ 1) = 0.50

Step-by-step explanation:

Given that:

The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.

Then; the probability of success = 7/34 = 0.20588

Using Binomial distribution to determine the probability; we have:

$P(X = x) = ^nC_x \ \beta^x \ (1 - \beta)^{n-x}$

where;

x = 0,1,2,...n and 0 < β < 1

and x represents the number of successes.

However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:

P(X ≥ 1) = 1 - P(X< 1)

P(X ≥ 1) = 1 - P(X =0)

$P(X \ge 1) = 1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]$

$P(X \ge 1) = 1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]$

P(X ≥ 1) = 1 - 0.50

P(X ≥ 1) = 0.50

4 0

3 years ago