PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER

You start a chain email and send it to six friends. The next day, each of your friends forwards the email to six people. The pro

Elina [12.6K]

Solution:

The following information about the chain email can be Written as follows:

1 Person Email →→6 Persons (E-mail) →6² Persons (E-mail)→6³ Persons (E-mail)+.........few Days

As you can see the number of email sent starting from person 1 to 6 persons to 36 persons to 216 persons, forms a Geometric pattern.

So, Sum of the Series = 1 + 6 + 36 + 216 + 1296 +......+ for n days

= $1+6^1+6^2+6^3+6^4 +6^5 +6^6+..........+ 6^n$

Formula for n terms of a geometric series

$\frac{a (r^n-1)}{r-1}$

For, Common ratio (r)≥1

Sum of above geometric series up to n terms $=\frac{6^n-1}{6-1}=\frac{6^n-1}{5}$

3 0

3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$