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MariettaO [177]
3 years ago
13

Help plz, will mark brainliest

Mathematics
2 answers:
kenny6666 [7]3 years ago
7 0

Answer:

d

Step-by-step explanation:

earnstyle [38]3 years ago
3 0

Answer:

option c.........it should be tan(20)=x/15.035

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A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with
AleksAgata [21]

Answer:

P (T) = 1/4

P ( T | F )    = 1/2  = P(F)

The events are not independent.

Step-by-step explanation:

Let F the event of picking the white ball first

P (F)= 1/2   ( picking the white ball first)

Let T be the event of getting the white ball twice,

P (T) = P( getting white ball) * P( getting white ball)

          =( 1/2)*(1/2)

           = 1/4

Here P(T∩F) = P(T) because the probability of getting the white balls is the same as probability of getting the white ball first  both the times.

P ( T | F ) = P (T∩F)/ P(F)

                = (1/4)/ (1/2)

                  = (1/2)

                    = 1/2  = P(F)

For the events to be independent the conditional probability P ( T | F )  must be equal to P(T).

Hence the events are not independent.

4 0
3 years ago
What is:<br> 2 x^4-x^2+3x+1 ÷ x^2+2x+2
Inessa [10]
Your answer is: 2x^6-x^4+5x^3+2x^2+1\x^2
6 0
3 years ago
Suppose that A = PDP-1 . Prove that det(A) = det(D)
daser333 [38]

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that  for two matrices A and B we have that:

det(AB) = det(A)*det(B) and det(A^{-1}) = \frac{1}{det(A)}. Now:

A = PDP^{-1}

det(A) = det(PDP^{-1})

det(A) = det(P)*det(D)*det(P^{-1})

det(A) = det(P)*det(D)*\frac{1}{det(P)}

det(A) = det(P)*\frac{1}{det(P)}*det(D)

det(A) = det(D).

5 0
3 years ago
Solve for k:<br> 3k + 5 = 23<br> EKKL
Musya8 [376]

Answer:

6

Step-by-step explanation:

3k + 5 = 23

3k = 23 - 5

    = 18

k = 18/3

  = 6

<em>Here's your answer :)</em>

3 0
3 years ago
Read 2 more answers
FILL IN THE BLANK
stepladder [879]
You’re answer is d (sum,sum).
7 0
3 years ago
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