If you apply a scale factor of 5 to the side measuring 8, what will the length of the new side be?

Mathematics

1 answer:

Leona [35]2 years ago

5 0

Answer:

Step-by-step explanation:

The scale factor of 5 to the side measuring 8 means that you multiply that side by the scale factor.

Basically, the equation would go like this:

Original Side Length: x

Scale Factor: y

New Side Length: z

xy=z

If we plug in our numbers, we get:

8*5=z

40=z

40

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Suppose that 35 people are divided in a random mannerinto two teams in such a way that one team contains10 people and the other

Alisiya [41]

To be precise, the size of your sample space is (2410)(2410). This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, (22)(22) (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are 24−2=2224−2=22 nondefective bulbs, and I must choose 88 of them, so that's (228)(228). Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: (22)(228)(22)(228)

P=(22)(228)(2410)P=(22)(228)(2410)

Or, since (22)=1(22)=1,

P=(228)(2410)P=(228)(2410)

Hope this helps!

7 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

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.

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

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

Now:

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