Scientific notation!

Which quadrilaterals always have opposite angles that are congruent

bazaltina [42]

Answer:

SQUARE

RHOMBUS

Step-by-step explanation:

A square is a parallelogram with four congruent sides and four right angles.

7 0

3 years ago

Jamal performed an experiment flipping a coin. He did 10 trials and then his arm got tired. He recorded his results in the table

AleksAgata [21]

Answer:

The error in his prediction is that he had too small of a sample size for the data to be reliable in how the trend would continue.

8 0

3 years ago

Read 2 more answers

En un huerto se siembra frijol. Si se pone una planta, produce 3kg de frijol. por cada planta extra que se ponga, todas las plan

Ad libitum [116K]

Answer:

y = -0.07x2 + 3.07x

Step-by-step explanation:

For each plant we add, there is a decrease of 70 grams for all plants from the inicial value of 3 kg of beans.

So, for 2 plants, we have (3 - 0.07) * 2

For 3 plants, we have (3 - 2*0.07) * 3

For x plants, we have (3 - (x-1)*0.07) * x

So we can model the final result of the amount of beans 'y' with the equation:

y = (3 - (x-1)*0.07) * x

y = (3 - 0.07x + 0.07) * x

y = (3.07 - 0.07x) * x

y = -0.07x2 + 3.07x

(In this equation we have a = -0.07, b = 3.07 and c = 0)

7 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

4 0

3 years ago

F equals to 2 f - 20

ICE Princess25 [194]

Answer:

20

Step-by-step explanation:

f = 2f - 20

f - 2f = - 20

- f = - 20

f = 20

7 0

3 years ago