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3 years ago

10

For his long distance phone service, Justin pays a $3 monthly fee plus 11 cents per minute. Last month, Justin's long distance b

ill was $20.16 . For how many minutes was Justin billed?

2 answers:

Setler79 [48]3 years ago

6 0

I'm sure it's 1.83 because you have to do 20.16 divided by 11

inna [77]3 years ago

4 0

$1.83. thats what i got when i did the math

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5/4 to 3/8 as a percent change

Digiron [165]

Well, what's the diffrence from 5/4 to 3/8?

$\bf \cfrac{5}{4}-\cfrac{3}{8}\qquad \stackrel{\textit{LCD is 8 clearly}}{\implies }\qquad \cfrac{10-3}{8}\implies \cfrac{7}{8}$

so, if we take 5/4 to be the 100%, what is 7/8 off of it in percentage anyway?

$\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{5}{4}&100\\\\
\frac{7}{8}&x
\end{array}\implies \cfrac{\quad \frac{5}{4}\quad }{\frac{7}{8}}=\cfrac{100}{x}\implies \cfrac{5}{4}\cdot \cfrac{8}{7}=\cfrac{100}{x}
\\\\\\
\cfrac{40}{28}=\cfrac{100}{x}\implies x=\cfrac{28\cdot 100}{40}$

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3 years ago

1. What is the length of the diagonal?<br> a²+ b²= c²<br> a 6 in. 'cin. I find c-<br> 6 in.

Aleksandr [31]

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I dont think you pasted the answer right... If you comment i can try and help you

Step-by-step explanation:

3 0

3 years ago

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In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

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3 years ago

Juan is 3 years younger than half his aunts age. If Juan is 27 years old, how old is his aunt?

Ugo [173]

Juan’s aunt is 57 years old

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2 years ago

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What are the solution(s) of y-4= 0?<br><br> X=-4 0r x= 4<br> X=-2 or x= 2<br> X=2<br> X=4

Alexxandr [17]

Answer: X=-2 or x= 2

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3 years ago

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