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ivanzaharov [21]
3 years ago
9

Solve 4x^2 = 70. Round to the nearest hundredth. The solutions are ± __________________

Mathematics
1 answer:
kicyunya [14]3 years ago
4 0
4x² = 70
x² = 70/4
x² = 17.5
x = ±√17.5
x = ±4.18

<span>The solutions are ±4.18</span>
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One of the wagers in roulette is to bet that the ball will stop on a number that is a multiple of 3. (Both 0 and 00 are not incl
goldenfox [79]

Answer:

The player's expectation is a loss of 20 cents.

Step-by-step explanation:

In a game of roulette, there are 38 slots where the ball can stop.

The slots are numbered as follows: {0, 00, 1, 2, 3 ..., 36}.

So, N = 38.

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So, n (S) = 12.

Let <em>X</em> = the ball stops on a number that is a multiple of 3

The probability that the ball stops on a number that is a multiple of 3 is:

P(X)=\frac{n (S)}{N}=\frac{12}{38}=0.32

It s provided that the player wins double the amount bet if the ball stops on a number that is a multiple of 3.

Compute the player's expectation as follows:

E(X)=\sum X\cdot P(X)

        =(\$10\times 0.32)+(-\$5\times (1-0.32))\\\\=\$3.20-\$3.40\\\\=-\$0.20

Thus, the player's expectation is a loss of 20 cents.

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For what values of a is the value of the binomial 2a-1 smaller than the value of the binomial 7-1.2a
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Step-by-step explanation:

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A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a ro
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Answer:

765 J

Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

W1 = (30,0)∫(0.2x .dx)

Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

m = 15/30 = 0.5

So,

F = 0.5x + 15

Thus,

W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

W2 = (0.5 × 30²)/2) + 15(30)

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W2 = 675 J

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

5 0
3 years ago
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