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3 years ago

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30. Joanne has a number of 10c coins and $1 coins in her purse, worth a total value of $6.60. She has one more than three times

as many 10c coins as $1 coins.
Let n be the number of $1 coins that Joanne has. Finish the equation describing Joanne's coins:
_*1.00+(_n+_) *_=6.60

1 answer:

olga nikolaevna [1]3 years ago

8 0

Answer:

oop

Step-by-step explanation:

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How do I solve this

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$\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\displaystyle\sum \limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf S_{20}=\displaystyle\sum \limits_{n=1}^{\stackrel{\stackrel{n}{\downarrow }}{20}}~\stackrel{\stackrel{a_1}{\downarrow }}{3}(\stackrel{\stackrel{r}{\downarrow }}{1.5})^{n-1}\implies S_{20}=3\left(\cfrac{1-1.5^{20}}{1-1.5} \right)\implies S_{20}=3\left(\cfrac{1-\stackrel{\approx}{3325.3}}{-0.5} \right) \\\\\\ S_{20}=3\left(\cfrac{-3324.3}{-0.5} \right)\implies S_{20}=3(6648.6)\implies S_{20}=19945.8$

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Mashutka [201]

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3 years ago

The line plots show the number of hours two different age groups of teenagers spent online the previous week.

nikklg [1K]

The range of the top plot, hours spent by 13-15 year olds, is found by subtracting the highest value, 34, and the lowest value, 21. 34-21=13.
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$\frac{21+24+26+27+27+28+28+28+31+34}{10}=\frac{274}{10}=27.4$

The range of the bottom plot, minutes spent online by 16-18 year olds, is found by subtracting the highest value, 36, and the lowest value, 12. 36-12=24.
The mean of the bottom plot is found by adding together all of the data points and dividing by 11 (the number of data points):
$\frac{12+21+27+28+28+28+28+30.5+34+35+36}{11}=\frac{307.5}{11} \approx27.95$

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mestny [16]

He saves $1253 per month.

Explanation: 65% of $3580 is $2327, meaning he spends $2327 per month. He has $1253 left over because $3580-$2327=$1253.

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