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dusya [7]
3 years ago
13

WILL GIVE BRAINLY!!Evaluate the following expression. (20 – 16)³ - 15

Mathematics
2 answers:
QveST [7]3 years ago
5 0

Answer:

49

Step-by-step explanation:

Vesnalui [34]3 years ago
4 0
The answer is 49 you first do the parentheses then you would get 4 and then you would do your exponet and get 64-15=49
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Which of the following could be the sides of a right triangle? Check all that apply.
VLD [36.1K]
A. & D. are the correct answers

To answer this, you square the first 2 numbers in the sequence, and then add them together. You then square the 3rd number and put it equal to the first numbers that you added together. If the two sides equal each other, it is correct!

Ex.:

10, 24, 26

10^2 + 24^2 = 26^2
100 + 576 = 676
676 = 676

18, 24, 30

18^2 + 24^2 = 30^2
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3 0
3 years ago
Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w
Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability  of getting heads is :

P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})

= \dfrac{P( \{Tails, Heads\})  \cap \{Tails, Heads,( Heads ,Tails)\})}{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) }{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) } {  {P( Tails, Heads) +P( Heads ,Tails)}}

=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}

=\dfrac{(1-p)*p}{2(1-p)*p}

=\dfrac{1}{2}

Thus; the probability of getting heads is \dfrac{1}{2} which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of  p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

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Answer:

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Step-by-step explanation:

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