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pentagon [3]
3 years ago
9

A coin is tossed and a number cube is rolled. What is P(heads, a number, a number less than 5)

Mathematics
2 answers:
KonstantinChe [14]3 years ago
7 0

That would be 1/2 * 4/6

= 1/2 * 2/3 = 1/3 (answer)

stepladder [879]3 years ago
5 0

In a coin, there are 2 sides - heads and tails. The probability of getting a head is \frac{1}{2}

In a number cube, there are 6 numbers. The probability of being a number is 1.

The 6 numbers in a number cube are: 1, 2, 3, 4, 5, 6. There are 4 numbers less than 5. The probability of being less than 5 is \frac{4}{6}.

Now multiply the 3 probabilities together:

\frac{1}{2} \times \frac{1}{1} \times \frac{4}{6} = \frac{4}{12} \rightarrow \frac{1}{3}

Your answer is A.

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3 years ago
Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet
Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\

Take the second derivative of the function, to check if it will given a minimum perimeter

\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)

Determine the critical points of the first derivative;

df/dw = 0

\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

3 0
2 years ago
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