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3 years ago

Independent Practice

Mathematics

2 answers:

mariarad [96]3 years ago

5 0

Answer:

B. (1, 0)

Step-by-step explanation:

Maru [420]3 years ago

4 0

9514 1404 393

Answer:

B. (1, 0)

Step-by-step explanation:

You can count grid squares on the graph to find the point that is 4 of the 8 units horizontally between the points, and 2 of the 4 units vertically between the points. That will be ...

C +(4, 2) = D -(4, 2) = (1, 0)

Or, you can use the midpoint formula:

M = (C +D)/2

M = ((-3, -2) +(5, 2))/2 = (-3+5, -2+2)/2 = (2, 0)/2

M = (1, 0)

The midpoint is (1, 0).

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2 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:

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

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liraira [26]

Answer:

Step-by-step explanation:

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