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1 year ago

Train A travels 30 miles in 20 minutes at a constant speed. Train B travels 20 miles in 15 minutes at a constant speed. Redo Whi

ch train is going faster? Circle your answer and show how you figured it out below, Train A Train B

Mathematics

1 answer:

krek1111 [17]1 year ago

3 0

$\begin{gathered} \text{Train A}\Rightarrow\frac{30\text{ miles}}{20\text{ minute}}=1.5\text{ miles/minute} \\ \text{ Train B}\Rightarrow\frac{20\text{ miles}}{15\text{ minute}}=1.34\text{ miles/minute} \\ \text{The train A is going faster.} \end{gathered}$

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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:

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

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