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3 years ago

What is the product of a number and 3?

2 answers:

8 0

The product is the result of multiplying, and whatever number is multiplied by 3 is the amount of times you add 3 to the answer.

Example:

3x7= 3+3+3+3+3+3+3=21

Mazyrski [523]3 years ago

5 0

A product of a number is the sum of a multiplication problem and 3=21

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ROUND YOUR ANSWER TO THE NEAREST TENTH

Darya [45]

Answer:

x = 16.5

Step-by-step explanation:

Use pythagorean theorem

a² + b² = c²

Plug in terms

7.1² + x² = 18²

Square each term

50.41 + x² = 324

Subtract a from c

x² = 273.59

Find square root

16.5

4 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

An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

Natasha2012 [34]

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

For this case the 9% confidence interval is given by:

$8.8104 \leq \mu \leq 11.1248$

We can calculate the mean with the following:

$\bar X = \frac{8.8104 +11.1248}{2}= 9.9676$

And we can find the margin of error with:

$ME= \frac{11.1248- 8.8104}{2}= 1.1572$

The margin of error for this case is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE$

And we can solve for the standard error:

$SE = \frac{ME}{t_{\alpha/2}}$

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

$SE = \frac{1.1572}{1.96}= 0.5904$

Now for the 98% confidence interval the significance is $\alpha=1-0.98= 0.02$ and $\alpha/2 = 0.01$ the critical value would be 2.326 and then the confidence interval would be: