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1 year ago

Which expressions are equivalent to RootIndex 3 StartRoot 128 EndRoot Superscript x? Select three correct answers.

Mathematics

2 answers:

vodka [1.7K]1 year ago

5 0

Answer:

<h3>- 128 Superscript StartFraction 3 Over x EndFraction </h3><h3>- (4RootIndex 3 StartRoot 2 EndRoot)x </h3><h3>- (4 (2 Superscript one-third Baseline) ) Superscript x</h3><h3>Step-by-step explanation:</h3>

Given the indicinal equation $(\sqrt[3]{128} )^{x}\\$

According to one of the law of indices,

$(\sqrt[a]{m} )^{b}\\= (\sqrt{m})^\frac{b}{a}$

Applying this law to the question;

$(\sqrt[3]{128} )^{x}\\ = {128} ^\frac{x}{3}\\ \\= (\sqrt[3]{64*2})^{x} \\ = (4\sqrt[3]{2})^{x} \\= (4(2^{1/3} )^{x} )$

The following are therefore true based on the following calculation

128 Superscript StartFraction 3 Over x EndFraction

(4RootIndex 3 StartRoot 2 EndRoot)x

(4 (2 Superscript one-third Baseline) ) Superscript x

Rzqust [24]1 year ago

4 0

Answer:

Options B, C, and D on edgenui.ty

Step-by-step explanation:

See the work down above ^^^

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Answer:

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We have the following data:

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- Probability that player V wins the 2nd set if he has won the 1st set:

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So, the probability that player V wins the first 2 sets is:

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Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

$p(2^c|1)=0.40$

So, the probabiity that player V winse the 1st set but loses the 2nd set is

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Also, we have:

- Probability that player V loses the 1st set:

$p(1^c)=0.50$

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

$p(2|1^c)=0.30$

So, the probability that she will lose the 1st set and win the 2nd set is:

$p(1^c2)=p(1^c)\cdot p(2|1^c)=(0.50)(0.30)=0.15$ (3)

Combined together (2) and (3), this means that the probability that player V wins exactly 1 set out of the first two sets is:

$p(1/2)=p(12^c)+p(1^c2)=0.20+0.15=0.35$ (4)

At this point, the probability that she will win the 3rd set is

$p(3)=0.45$

This means that the overall probability that she will win the 3rd set if she won exacty 1 of the first 2 sets is:

$p(1/2,3) = p(1/2)\cdot p(3)=(0.35)(0.45)=0.16$ (5)

So, the overall probability that player V will win a match against player M is the sum of (1) and (5):

$p(W)=p(12)+p(1/2,3)=0.30+0.16=0.46$

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