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4 years ago

What is the area of a triangle that has a base of 15 1/4 in. And a height of 18 in.

Mathematics

1 answer:

Minchanka [31]4 years ago

8 0

Answer:

144

Step-by-step explanation:

Use the triangle formula base times height times 1/2 (1/2xbxh) so 15.25 (15 1/4) times 18 = 288. Then you can divide 288 by 2 or multiply it by 1/2 you get the same answer regardless which the final answer is 144.

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The figure is the net for a rectangular prism.

oksano4ka [1.4K]

Answer:

$SA=544\ cm^2$

Step-by-step explanation:

we know that

The surface area of a rectangular prism is equal to the area of its six rectangular faces

using the net

The surface area is equal to

$SA=2[(16)(6)]+2[(16)(8)]+2[(8)(6)]$

$SA=192+256+96$

$SA=544\ cm^2$

3 0

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!

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

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

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.

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

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