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charle [14.2K]
4 years ago
8

What is the equivalent decimal for -1.25

Mathematics
2 answers:
Sphinxa [80]4 years ago
8 0
1 1/4 why? because:

25% is equal to 1/4 and the 1 ones is equal to 1 whole

Hope this helps! :) 
Hope you get it right if you do please put myself <span>brainliest and Good luck! </span>
Tanzania [10]4 years ago
4 0
I honestly do not know what to say here: it's already in decimal form. Please, if you made a mistake while typing in the problem, tell me.
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Is 169 a whole number,Interger,Rational, Irrational,or Real?
sergeinik [125]

Answer:

A whole number, an integer, a rational number and a real number.

Step-by-step explanation:

It is a whole number, which makes it an integer, which makes it rational, which makes it real.

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3 years ago
What is 275 equals n multiplied by n by 216 as an equation
loris [4]

Answer:

275=n× n by 216

Step-by-step explanation:

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!

<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
kelly lost 14.6 pounds over 4 weeks. if she lost the same each week, what was the weekly change in her weight.
Alisiya [41]

14.6 lbs/4 weeks = 3.65 lbs /week

she lost  3.65 pounds each week

6 0
3 years ago
Read 2 more answers
Find f(-3) it f(x) =x^2
xeze [42]

Answer:

9

Step-by-step explanation:Replace the variable  

x  with  − 3  in the expression.  f ( − 3 )  =  ( − 3 ) 2  Simplify  (− 3 ) 2 .  Remove parentheses.  ( − 3 ) 2  Raise  − 3  to the power of  2 . 9

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