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FromTheMoon [43]
3 years ago
13

for independence day celebration 25000 flags were distributed to 70 schools .if 40 shcool receive 250 each and the remainder dis

tributed equally among the other schools calculate the number of flags receives
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

500 flags each

Step-by-step explanation:

If 40 schools receive 250 each, then in total, that's 10,000

25,000 - 10,000 = 15,000.   There are 30 schools remaining, so we divide.  15,000 / 30 = 500, so...

40 schools get 250 flags each, and the other 30 schools get 500 flags each.

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The answer to this is: 8
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a construction company charges $15 per hour for debris removal , plus a one -time fee for the use of the trash dumpster.the tota
ser-zykov [4K]

Let h be the numbers of hours, f be the one-time fee and c the cost charged. The equation is

c=15h+f

Since you pay $15 per four, plus the fee. We can solve this equation for the fee:

f=c-15h

If we plug c=195 and h=9, we have

f=195-15\cdot 9=195-135=60

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Espen predicted that he would sell 32 baseball caps but he actually sold 29 baseball caps which expression would find the percen
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Answer:

B

Step-by-step explanation:

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6 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
Do problem 6 show work i will give brainliest if ur correct.
aliina [53]

Step-by-step explanation:

they forget the last step which is to take the square root of the sum

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305 = c²

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7 0
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