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Naddik [55]
3 years ago
9

the ratio of the number of adults to the number of students at the prom has to be 1:10. Last year there were 477 more students t

han adults at the prom. If the school is expecting the same attendance this year, how many adults have to attend the prom?
Mathematics
1 answer:
Alexeev081 [22]3 years ago
6 0
Ratio:\ \ \frac{1}{10}\\\\x-number\ of\ adults\\477+x-number\ of\ students\\\\If\ ratio\ is\ \frac{1}{10}, for\ 1\ adult\ is\ 10\ students.\ So\ for\ x\ adults\ is\ 10x\\students.\\\\10x=477+x\ \ \ \ \ \ |substract\ x\\10x-x=477\\9x=477\ \ \ \ |:9\\x=53\\\\10x=10\cdot53=530\\\\There\ were\ 53\ adults\ and\ 530\ students.
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If a coin is flipped 20 times and a head comes up 7 times, what is the relative frequency of a head coming up?
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Relative frequency is the absolute frequency (i.e. the number of times of the wanted result) divided by the total  number of events.

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Number of times the coin was flipped: 20
Numbef of times head came up: 7

Relative frequency: 7 / 20 = 0.35, which is the option B.
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Five less than one third of a number is two more than half of the same number. Find the number.
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How many terms of the series of - 3+0+3+6+9+...are needed to give a sum of 105?​
dolphi86 [110]

Answer:

10

Step-by-step explanation:

Remember that the formula for the sum of an arithmetic series is:

S=\frac{k}{2}(a+x_k)

Where k is the number of terms, a is the initial term, and x_k is the last term of the series.

We essentially want to find k, the number of terms, given that the sum S is equal to 105. So, substitute 105 into our equation:

105=\frac{k}{2}(a+x_k)

To do so, we need to final term x_k. We don't know what it is yet, but that doesn't matter. All we need to do is to write it in terms of k. First, remember that the standard form for the explicit formula of an arithmetic sequence is:

x_n=a+d(n-1)

Where a is the first term, d is the common difference, and n is the nth term.

From our sequence, we can see that the first term is -3.

Also, we can determine that our common difference is +3, since each subsequent term is 3 <em>more</em> than the previous one. -3+3 is 0, 0+3 is 3, 3+3 is 6, and so on.

Therefore, our explicit formula is:

x_n=-3+3(n-1)

Therefore, our final term, x_k, will be if we substitute k for n. So, we can acquire the equation:

x_k=-3+3(k-1)

Now that we know what x_k is, we can substitute that into our original equation:

105=\frac{k}{2}(a+x_k)

Substitute the equation into x_k. Also, let's substitute -3 (our first term) for a. So:

105=\frac{k}{2}(-3+(-3+3(k-1)))

And now, all we have to do is to solve for k.

First, distribute the 3:

105=\frac{k}{2}(-3+(-3+3k-3))

Add within the parentheses:

105=\frac{k}{2}(3k-9)

Multiply both sides by 2. This removes the fraction on the right:

210=k(3k-9)

Distribute. We will get a quadratic:

210=3k^2-9k

So, let's solve for k. Let's divide everything by 3:

70=k^2-3k

Subtract 70 from both sides:

0=k^2-3k-70

Factor. We can use -10 and 7. So:

0=(k-10)(k+7)

Zero Product Property:

k-10=0\text{ or } k+7=0

Solve for k for each equation:

k=10\text{ or } k=-7

-7 doesn't make sense (we can't have -7 terms). Remove that solution. So, we are left with:

k=10

Therefore, the number of terms we have in our series for our sum to be 105 is 10.

And we're done!

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3 years ago
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