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In-s [12.5K]
2 years ago
15

The volume of the cone shown is 5 cubic inches. What is the height of a cone with the same base diameter but a volume of 10 cubi

c inches?
Mathematics
1 answer:
Alisiya [41]2 years ago
3 0

Answer:

The height of the big cone is double the one in the small cone

h2 = 2h1

Step-by-step explanation:

Given that:

  • The volume of the small cone: 5 cubic inches
  • The volume of the big cone: 10 cubic inches

As we know, the volume of a cone is as following:

V = (1/3)*area of the base*height

If the base diameter are unchanged => area of the base of the two cones are  unchanged and from the given information, the volume of the big cone is double the volume of the small cone. So the height of the big cone is double the one in the small cone

<=> h2 = 2h1

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<span>The area of the square is d^2. The area of the circle sandbox is πr^2=πd^2/4. So the area of play area only is equal to square area minus circle area which is d^2-πd^2/4.</span>
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3 years ago
For which equation is x = 9 a solution? A. 3x – 7 = –34 B. x – 9 = 1 C. 2003-05-02-00-00_files/i0020000.jpg D. 4x + 5 = 41
Monica [59]
X=9


A) 3x-7=34
 3x= 34+7
 3x= 41
 x= 41/3
 

B) x-9=1
x=1+9
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C) dude what is this


D) 4x+5=41
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3 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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What is the perimeter of the rectangle? one side is 6in. and the A=30 sq in.
monitta
The perimeter of a rectangle can be found by dividing the area by one side.
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3 years ago
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Given teo circles whose radii are 9" and 24", the ratio of thuer circumfrence is?
saul85 [17]

Answer:

The ratio is 3:8

Step-by-step explanation:

The circumference of a circle can be calculated using the formula;

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C = 2 * pi * 9 = 18 pi inches

For 24 inches;

C = 2 * pi * 24 = 24 pi inches

The ratio is thus;

18 pi : 48 pi

divide through by 6 pi

= 3: 8

5 0
2 years ago
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