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3 years ago

Help me with this pleaseeeeeeeeeee

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

C (-3,5)

Step-by-step explanation:

hope this helps!

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xxMikexx [17]

Answer:

Step-by-step explanation:

The quantity x minus 2y is (x - 2y), then we add 3.

4 0

4 years ago

ASAP PLS ANSWER HELP

castortr0y [4]

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Answer: Answer #1 in top left corner

Explanation:

Only have 15 sites represented

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

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4 0

3 years ago

Read 2 more answers

ALG II NEED HELP QUICK! Will mark the brainliest answer

lbvjy [14]

I think it’s the second one because The formula would be
x=
2a
−b±
b
2
−4ac

4 0

3 years ago

I've done the sketch but am unsure as what my next steps should be?

Grace [21]

Answer:

you gotta see whats up with it

Step-by-step explanation:

7 0

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