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WARRIOR [948]
3 years ago
14

Simplify

rac{3}{8} } " alt=" \sqrt{ \frac{3}{8} } " align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Pie3 years ago
3 0

Answer:

B (√6)/4

Step-by-step explanation:

The smallest multiplier that will make the denominator of the fraction into a perfect square is 2, so you have ...

\displaystyle\sqrt{\frac{3}{8}}=\sqrt{\frac{3\cdot 2}{8\cdot 2}}\\\\=\sqrt{\frac{6}{16}}=\frac{\sqrt{6}}{\sqrt{16}}\\\\=\frac{\sqrt{6}}{4}

_____

Answer choice D is also a correct rationalization of the denominator, but is not simplified as far as it can be. √24 = 2√6, so a factor of 2 can be cancelled from numerator and denominator, giving answer choice B.

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A bank in the Bay area is considering a training program for its staff. The probability that a new training program will increas
WITCHER [35]

Answer:

P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68

See explanation below.

Step-by-step explanation:

For this case we define first some notation:

A= A new training program will increase customer satisfaction ratings

B= The training program can be kept within the original budget allocation

And for these two events we have defined the following probabilities

P(A) = 0.8, P(B) = 0.2

We are assuming that the two events are independent so then we have the following propert:

P(A \cap B ) = P(A) * P(B)

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

P(B' \cup A')

And using the De Morgan laws we know that:

(A \cap B)' = A' \cup B'

So then we can write the probability like this:

P(B' \cup A') = P((A \cap B)')

And using the complement rule we can do this:

P(B' \cup A') = P((A \cap B)')= 1-P(A \cap B)

Since A and B are independent we have:

P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32

And then our final answer would be:

P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68

5 0
3 years ago
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