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2 years ago

Simplify (2/9)^3 A.6/9 B.8/729 C.2/720

Mathematics

1 answer:

saw5 [17]2 years ago

8 0

Answer:

$\frac{8}{729}$

Explanation:

Before we begin, remember that:

( $\frac{a}{b}$ )ⁿ = $\frac{a^n}{b^n}$

This means that power is distributed in case of division

The given is:

$(\frac{2}{9} )^3$

Applying the above rule, we would get:

$(\frac{2}{9} )^3$ = $\frac{2^3}{9^3} = \frac{2*2*2}{9*9*9} = \frac{8}{729}$

Hope this helps :)

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The bookstore has 10 new copies and 10 used copies in stock. If 15 people come in one by one to purchase this text, what is the

stellarik [79]

Answer:

The probability that all 15 will get the type of book they want from current stock is 0.4838.

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other 70% want a used copy. Consider randomly selecting 15 purchasers.

The bookstore has 10 new copies and 10 used copies in stock.

If 15 people come in one by one to purchase this text, what is the probability that all 15 will get the type of book they want from current stock?

Here is my answer:

Given:

n= 15

P (want used copy) = 0.7

P (want new copy) = 0.3

Let X = the number who want a new copy.

All the 15 students get their desired copy, then this can happen if at most 10 want to buy new copy and at least 5 wants to buy used copy.

So compute the probability of (5 ≤ X ≤ 10) as follows:

P (5 ≤ X ≤ 10) = P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)

= $C^{5} _{15}$ * $0.3^{5}$ * $0.7^{10}$ + $C^{6} _{15} *0.3^{6} *0.7^{9}$ + $C^{7} _{15} *0.3^{7} *0.7^{8}$ + $C^{8} _{15} *0.3^{8} *0.7^{7}$ + $C^{9} _{15} *0.3^{9} *0.7^{6}$ + $C^{10} _{15} *0.3^{10} *0.7^{5}$

= 0.4838

So the probability that all 15 will get the type of book they want from current stock is 0.4838.

Hope it will find you well.

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2 years ago