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2 years ago

How many times can henry expect to roll a 4 if he rolls a dice 120 times

Mathematics

1 answer:

olga2289 [7]2 years ago

5 0

Answer: Varies

Step-by-step explanation:

There's a 1 / 6 chance each time.

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What is the value of 3.45 + 2.6 + 32.007?

bulgar [2K]

38.057 is the total

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2 years ago

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4. Using the point-slope formula, what is the equation of the line that passes through the points (1, 5) and (0, 0)?

Mazyrski [523]

<span>Answer: y=5x or 5x-y=0
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3 years ago

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Suppose that E and F are two events and that N (E and F) = 260 and N(E) = 630. What is P(F|E)?

Sveta_85 [38]

Answer: 0.4127

Step-by-step explanation:

The probability formula to find p(F/E):

p(F/E)= N(E n F) / N(E)

p(F/E)= 260 / 630

p(F/E) = 0.4127

5 0

3 years ago

I have to transfer this graph into a vertex form equation, but I'm not sure how to.

tino4ka555 [31]

Answer: y = -2/3(x-3)^2 + 0 or y = -2/3(x-3)^2

Step-by-step explanation:

vertex form is y=a(x-h)^2 + k

here we can see the vertex is (3,0) which is (x,y). Or (h,k) in this case.

so to plug that into vertex form, we now have y=a(x-3)^2 + 0. or just y=a(x-3)^2.

now we need to find "a" which is the leading coefficient. to do that we can plug in the (6,-6) for the x and y parts of the above equation. so we'd have

-6=a(6-3)^2. which goes to -6=a(2)^2 which is -6=4a. divide each side by 4 to get a = -2/3. plug this in for a

the final equation would be y = -2/3(x-3)^2 + 0 or y = -2/3(x-3)^2

8 0

2 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

3 years ago