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Kay [80]
3 years ago
9

Can you help me with three questions? -2x + 3 -x + 4 -x - 19

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0

-2x=-3

x=1.5

-x=-4

x=4

-x=19

x=-19

Hope this helps?

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Which real-world story is represented by the addition problem (–2 1/2) + (–3 1/4) = –5 3/4?
allochka39001 [22]

Answer:

A

Step-by-step explanation:

A is the correct answer because the equation matches option A.

(-2 1/2) + (-3 1/4)

The minus sign in -2 1/2 shows that the submarine descended and then descended again with the addition sign. Though the addition sign can mean that it ascended its incorrect.

With the equation in hand, we can prove that option A is correct.

If the submarine descended 2 1/2 miles then here must be a minus sign next to the 2 1/2. But if it ascended then there would've been an addition sign to show that it ascended. The other value - 3 1/4 also shows that it ascended with the help of the addition sign.

Ascends - values increase from smallest to highest.

Descends - values decrease from highest to smallest.

4 0
3 years ago
0.23 or 2.3% which is greater?
lisabon 2012 [21]

Answer:

0.23

Step-by-step explanation:

2.3% = 0.023

So 0.23*100=23% therefore larger than 2.3%


Any questions feel free to ask. Thanks

8 0
3 years ago
Could I get some help? Its simplifying square roots of negative numbers.​
Ainat [17]

Answer:

The square root of -100 is 10i.

Step-by-step explanation:

8 0
3 years ago
WILL GIVE BRAINLEST!! Two cards are drawn at random from a standard deck of 52 cards. What is the probability that they have the
Natali5045456 [20]

Answer:

It is unlikely that they will be the same rank 2/52 chance witch is a 0.03% chance

Step-by-step explanation:

4 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!

<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

4 0
3 years ago
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