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alexandr1967 [171]
3 years ago
5

Mario has a business selling muffins. Let x be the price of a muffin. Then, the profit P for Mario’s business is given by p(x)=-

100x2+350x-150 Choose the inequality that shows the business will make a positive profit.
Mathematics
2 answers:
Alex3 years ago
6 0

The first question's answer is:

0<-100x^2+350x-150

The second question "Look at the factorization shown below.

0 < –100x2 + 350x – 150

0 < –50(2x2 – 7x + 3)

0 < –50(2x – 1)(x – 3)

Select the range that Mario can choose from to price his muffins and make a positive profit.

The answer is:

$0.50 < x < $3.00

Rufina [12.5K]3 years ago
5 0
To make a positive profit p(x)>0 we need to make:
-100 x^{2} +350x-150 \ \textgreater \  0

Now we solve this for x:
-2 x^{2} +7x-3 \ \textgreater \ 0

We have:
a = -2
b = 7
c = -3

We will use formula for quadratic equation:
x_{1} =  \frac{-b+ \sqrt{ b^{2}-4ac } }{2a}  \\  \\  x_{2} =  \frac{-b- \sqrt{ b^{2}-4ac } }{2a}  \\  \\  x_{1} =  \frac{-7+ \sqrt{ 49-24 } }{-4} = \frac{-7+5 }{-4} = \frac{-2 }{-4} = \frac{1}{2}  \\  \\  x_{2} =  \frac{-7- \sqrt{ 49-24 } }{-4}  = \frac{-7-5 }{-4} = \frac{-12 }{-4} = 3

We got two solutions. One is fraction other is whole number. We will not consider fraction because the amount of muffins sold must be whole number. So our solution is:
x>3
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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

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Answer:

1)B

2)C

3)B

4)A

5)C

Step-by-step explanation:

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Problem 23
A = 40
B = 3x+1
C = 3x+1
B and C are congruent, so they have the same angle measure

The three angles of any triangle add to 180 degrees
A+B+C = 180
40+(3x+1)+(3x+1) = 180
6x+42 = 180
6x = 180-42
6x = 138
x = 138/6
x = 23

Answer: B) 23

============================================================

Problem 24

AB is the midsegment parallel to XY
So AB is half that of XY
AB = (1/2)*XY
AB = (1/2)*28
AB = 14

Answer: B) 14

============================================================

Problem 25

Note the ratio of the given rectangle's sides: 7/12 = 0.583 roughly

Divide the answer choices in a similar fashion
A) 5/7 = 0.714
B) 14/36 = 0.389
C) 21/36 = 0.583
D) 21/48 = 0.4375

We get the same result in choice C as we did in the original rectangle

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Answer:

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