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saveliy_v [14]
3 years ago
13

A thermometer gives a reading of 26 degrees Celsius. Use the formula c =5/9(f-32). Write the inverse function and use it to find

the equivalent temperature in Fahrenheit
Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

The answer to your question is F = 78.8

Step-by-step explanation:

Equation

                           C = \frac{5}{9} (F - 32)

Multiply both sides by 9

                           9C = \frac{5(9)}{9} (F - 32)

Simplify

                           9C = 5(F - 32)

Divide both sides by 5

                           \frac{9}{5} C = \frac{5}{5} (F - 32)

Simplify

                           \frac{9}{5} C = F -32

Add 32 in both sides

                           \frac{9}{5} C + 32 = F - 32 + 32

Simplify (Inverse)

                           \frac{9}{5} C + 32 = F

Substitute to find F

                           \frac{9}{5} (26) + 32 = F

Simplify

                                          F = 46.8 + 32

Result

                                          F = 78.8    

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Four times the amount of money in my bank account plus $200 is still two
ioda

Answer:

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Step-by-step explanation:

So, I like to extract the information we know and form it into an equation.

So we know that after multiplying what you have in the bank by 4, then add 200 dollars you get two thousand dollars less than a fourth of what's in your dad's bank account.

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Before we write the equation let's find what two thousand less than a fourth of 20,000 is.

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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>

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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}

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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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