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3 years ago

15

Solve 3x - x + 2 = 12

1 answer:

Anna11 [10]3 years ago

7 0

Remember you can do anything to an equation as long as you do it to both sides (except divide by 0)

add like terms
3x-x=3x-1x=2x

so
3x-x+2=12
2x+2=12
subtract 2 from both sides
2x=10
divide both sides by 2
x=5

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Angles α and β are the two acute angles in a right triangle. Use the relationship between sine and cosine to find the value of β

Dimas [21]

Answer:

From given relation the value of β is 37.5°

Step-by-step explanation:

Given as :

α and β are two acute angles of right triangle

Acute angle have measure less than 90°

Now given as :

$sin(\frac{x}{2} + 2x)$ = $cos(2x +\frac{3x}{2})$

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SO, $(90° - (\frac{x}{2}+2x))$ = $2x+\frac{3x}{2}$

Or, 90° = $2x+\frac{3x}{2}$ + $\frac{x}{2}+2x$

or, 90° = $\frac{4x}{2}$ + 4x

Or, 90° = $\frac{12x}{2}$

So, x = $\frac{90}{6}$ = 15°

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So, $sin(\frac{x}{2} + 2x)$ = sin $\frac{75}{2}$

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3 years ago

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Dafna1 [17]

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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:

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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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3 years ago