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uranmaximum [27]
3 years ago
15

Ten people from a local youth group went to black hills whitewater rafting tour company for a one day rafting trip. the group pa

id $425.
Write an equation in slope-intercept form to find the total cost C for p people.

How much would it cost for 15 people?
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
6 0

42.5p=c

You get you get the number 42.5 since you know 10 people paid 425, you would divide 425 by 10 to know how much every person paid.

M is representing the amount of money each person paid

(M)10=425
M=425/10
M=42.50 <- since it’s about money

Each person paid $42.50


The cost for 15 people is $637.50.

42.5p=c
42.5(15)=637.5
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Step-by-step explanation:

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Step-by-step explanation:

Z-tests are statistical calculations that can be used to compare the population mean to a sample mean The z-score is used to tellsbhow far in standard deviations a data point is from the mean of the data set. z-test compares a sample to a defined population and is typically used for dealing with problems relating to large samples (n > 30). Z-tests can also be used to test a hypothesis. Z-test is most useful when the standard deviation is known.

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3 years ago
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Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

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\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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