which expression have a value of 16/81? check all that apply. (2/3)^4, (16/3)^4, (4/81)^2, and (4/9)^2

2 answers:

6 0

$Use\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\left(\dfrac{2}{3}\right)^4=\dfrac{2^4}{3^4}=\dfrac{16}{81}\ :)\\\\\left(\dfrac{16}{3}\right)^4=\dfrac{16^4}{3^4}=\dfrac{16^4}{81}\neq\dfrac{16}{81}\ :(\\\\\left(\dfrac{4}{81}\right)^2=\dfrac{4^2}{81^2}=\dfrac{16}{81^2}\neq\dfrac{16}{81}\ :(\\\\\left(\dfrac{4}{9}\right)^2=\dfrac{4^2}{9^2}=\dfrac{16}{81}\ :)$

chubhunter [2.5K]3 years ago

5 0

Answer:

(2/3)⁴, (4/9)² apply to 16/81

Step-by-step explanation:

16/81

8 * 2/27 * 3

4 * 2 * 2/9 * 3 * 3

2 * 2 * 2 * 2/3 * 3 * 3 * 3

You might be interested in

On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre

Zarrin [17]

Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

$P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}$

$P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514$

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

$P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}$