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Ostrovityanka [42]
3 years ago
8

Billie’s monthly cell phone bill includes the cost of cell phone service, 15% tax on the cost of the cell phone service, and a $

9.25 data fee. Billie’s monthly cell phone bill totals $61.
What is the cost of the cell phone service before tax is added?
Mathematics
1 answer:
Mashcka [7]3 years ago
4 0
<em>It would be $46 because when you add the 15% tax to it, the total is $61. </em>
<em>$61-15%=$46. To check $46+15%=61! Hope i helped! :-)</em>
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Help please:Suppose that x and y inversely, and x=10 when y=8 write the function that models the inverse variation
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SOLUTION

From the question, we are told that y varies inversely as x

This is mathematically written as

\begin{gathered} y\propto\frac{1}{x} \\ \propto\text{ is a sign of proportionality } \end{gathered}

Now, we will remove the proportionality sign and replace it with equal to sign =

If we do this, we will intoduce a constant k

\begin{gathered} y\propto\frac{1}{x} \\ y=k\times\frac{1}{x} \\ y=\frac{k}{x} \end{gathered}

So we have the formula

y=\frac{k}{x}

We will substitute the values of x for 10 and y for 8 into the formula to get k, we have

\begin{gathered} y=\frac{k}{x} \\ 8=\frac{k}{10} \\ k=8\times10 \\ k=80 \end{gathered}

Now, we will substitute k for 80 back into the formula to get the inverse function, we have

\begin{gathered} y=\frac{k}{x} \\ y=\frac{80}{x} \end{gathered}

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What is another way to write the number 271
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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

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