All categories

3 years ago

Help again!!!!!!!!!!!!!!!

Mathematics

1 answer:

balandron [24]3 years ago

7 0

Answer:

4.5:7.25

Step-by-step explanation:

she ran 7.25 miles so ig that would be it but im not 100% sure

You might be interested in

Suppose a marketing company computed a 94% confidence interval for the true proportion of customers who click on ads on their sm

Ksivusya [100]

Answer:

d. There is a 98% chance that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62.

Step-by-step explanation:

Confidence interval:

x% confidence

Of a sample

Between a and b.

Interpretation: We are x% sure(or there is a x% probability/chance) that the population mean is between a and b.

In this question:

I suppose(due to the options) there was a small typing mistake, and we have a 98% confidence interval between 0.56 and 0.62.

Interpreation: We are 98% sure, or there is a 98% chance, that the true population proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. Option d.

3 0

3 years ago

A total of 50,000 contestants are chosen for a survey. A

ankoles [38]

I am pretty sure it is linear decrease,

4 0

3 years ago

What is the value of s?

Leya [2.2K]

Answer: a value is like how much is the pice or like counting money

Step-by-step explanation:

8 0

3 years ago

Give an example of two monomials with a quotient of -3n²/m

natima [27]

Hmmmm I’ll have to calculate this

3 0

3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

1 year ago