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3 years ago

What are the two events that are compared in the second and third

Mathematics

1 answer:

Advocard [28]3 years ago

4 0

Answer: second and third what?

Step-by-step explanation:

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A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the

Arturiano [62]

you do 15.0 x 4.0 and that would give you the answer of 60.

the bridge is 60m above the water.

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3 years ago

Please help me with this homework

Akimi4 [234]

Answer:

Additive Identity I think, I'm not sure

Step-by-step explanation:

Tell me in the comments if I'm wrong.

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2 years ago

James is 60 years old. His age is 8 years more than twice Maggie's age. How old is Maggie?

kow [346]

Answer:26

Step-by-step explanation:

60-8=52

52•2=26

4 0

3 years ago

Read 2 more answers

The set {6, 7, 8, 9, 10} is part of the solution set for which inequality?

slava [35]

7, is the part of the solution set because of the number it has inside of it. 8 years old worth of

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3 years ago

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