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slavikrds [6]
3 years ago
7

If sin(x) = cos(y) for acute angles x and y, how are the angles related?

Mathematics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

b. complementary

Step-by-step explanation:

-Complementary angles are angles that add up to 90°.

-These are usually the two acute angles in the right triangle.

#To verify, lets take the two angles 30° and 60°:

Cos \ 60\textdegree=0.5\\\\Sin \ 30\textdegree=0.5\\\\\therefore Sin \ 30\textdegree=Cos \ 60 \textdegree=0.5

#We can reverse as:

Sin \ 60\textdegree=0.86603\\\\Cos \ 30\textdegree=0.86603\\\\\therefore Sin \ 60\textdegree=Cos \ 30\textdegree=0.86603

Hence, two angles are said to be complimentary if they sum up to 90°.

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A random sample of 38 wheel chair users were asked whether they preferred cushion type A or B, and 28 of them preferred type A w
-BARSIC- [3]

Answer:

The statement that cushion A is twice as popular as cushion B cannot be verified

Step-by-step explanation:

From the question we are told that:

Sample size n=38

Type a size A X_a=28

Type a size B X_b=10  

Generally the probability of choosing cushion A P(a) is mathematically given by

P(a)=\frac{28}{38}

P(a)=0.73

Generally the equation for A to be twice as popular as B is mathematically given by

P(b)+2P(b)=3P

Therefore Hypothesis

Null H_0: p \leq \frac{2P}{3P} \\Altenative H_A:p>\frac{2P}{3P}

Generally the equation normal approx of p value is mathematically given by

z=\frac{x-np_0-0.5}{\sqrt{np_0(1-p_0)} }

z=\frac{28-(38*2/3)_0-0.5}{\sqrt{38*2/3*1/3} }

z=0.75

Therefore from distribution table

Pvalue=1-\theta (0.75)

Pvalue=0.227

Therefore there is no sufficient evidence to disagree with  the Null hypothesis H_0

Therefore the statement that cushion A is twice as popular as cushion B cannot be verified

7 0
3 years ago
A. 2x4 + 21x2 +3x -2<br>B. 2x4 + 9x2 -2<br>C. 3x2 + 6x+ 1 <br>D. 2x4 + 9x3 +12x2 + 3x -2 <br> ​
Inga [223]
You need to foil, The answer is D
6 0
3 years ago
How many counterexamples do you need to prove a conjecture to be false
Hoochie [10]

Answer:

one , i guess....

because the conjecture is only true if it satisfies all the cases.

3 0
3 years ago
Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of
steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

  • From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack)  P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
  • A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

  • WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the  probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

  • WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a  king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been  removed.

  • WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick  a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

  • The probability of drawing the second queen is also  4/52 and the third is 4/52.
  • We multiply these three individual probabilities together to get P(QQQ) =
  • P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
  • Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
5 0
3 years ago
Please help me. Explain this and give answer
Trava [24]

Answer:

It's B) 60

Step-by-step explanation: Just look at the similar one above T.

8 0
3 years ago
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