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4 years ago

9

Suppose that you ask three friends to go to the mall. each one has a 0.80 chance of saying yes. what is the probability that all

three friends will say they can go to the mall with you?

2 answers:

blondinia [14]4 years ago

5 0

The probability is 0.512

Lyrx [107]4 years ago

5 0

Answer:

0.512

Step-by-step explanation:

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0.5s +1=7 + 4.5s<br> S =

kvv77 [185]

Answer:

$s = -\frac{3}{2}$

Step-by-step explanation:

Solve for the value of $s$ :

$0.5s + 1 = 7 + 4.5s$

-Combine both $0.5s$ and $4.5s$ by subtracting $0.5s$ by $4.5s$ :

$0.5s + 1 - 4.5s = 7 + 4.5s - 4.5s$

$-4s + 1 = 7$

-Subtract $1$ to both sides:

$-4s + 1 - 1 = 7 - 1$

$-4s = 6$

-Divide both sides by $-4$ :

$\frac{-4s}{-4} = \frac{6}{-4}$

$s = -\frac{3}{2}$

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3 years ago

Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?

Mademuasel [1]

Answer:- a.The given expression is equivalent to $\frac{x^{10}y^{14}}{729}$

Given expression:- $[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}$

$=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}$

$=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}$

$=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}$

$=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}$

$=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}$

Thus a. is the right answer.

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