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2 years ago

14

I’m not sure what this question is asking or how to solve, please help asap

2 answers:

svet-max [94.6K]2 years ago

8 0

Answer:

divide 15 from A

Step-by-step explanation:

bazaltina [42]2 years ago

7 0

Answer:

a - 15

b - 7

c - 4

Step-by-step explanation:

To answer this, you need to know how to rewrite the form of the expression: $\sqrt[c]{a^{b} }$ . To rewrite this, use the following formula: $a^{\frac{b}{c} }$ . In your problem, the expression is: $\sqrt[4]{15^{7} }$ . Rewite this using the formula. In your problem the answer to this would be: $15^{\frac{7}{4} }$ . Break this down to understand that:

a - 15

b - 7

c - 4

Hope it helps!

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2 years ago

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insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

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where the last fact is due to the law of total probability:

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$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

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By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

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We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

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$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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