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exis [7]
2 years ago
12

Triangle ABC is a right triangle and sin(53o) = StartFraction 4 Over x EndFraction. Solve for x and round to the nearest whole n

umber.
Triangle A B C is shown. Angle A C B is a right angle and angle B A C is 53 degrees. The length of B C is 4 centimeters, the length of A C is y, and the length of hypotenuse A B is x.

Which equation correctly uses the value of x to represent the cosine of angle A?

cos(53o) = StartFraction 4 Over x EndFraction
cos(53o) = StartFraction y Over 5 EndFraction
cos(53o) = StartFraction x Over 4 EndFraction
cos(53o) = StartFraction 5 Over y EndFraction

Mathematics
1 answer:
zmey [24]2 years ago
4 0

Answer:

cos(53°)=StartFraction y Over 5 EndFraction

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the value of x

we know that

sin(53\°)=\frac{4}{x}

Solve for x

x=\frac{4}{sin(53\°)}

x=5

step 2

Find the cosine of angle A

The cosine of angle A is equal to divide the adjacent side to angle A (AC) by the hypotenuse (AB)

cos(53\°)=\frac{y}{x}

substitute the value of x

cos(53\°)=\frac{y}{5}

so

cos(53°)=StartFraction y Over 5 EndFraction

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One: B

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Step-by-step explanation:

Problem One

There are only two numbers in the sample of 40 that are under 26. Both are 25. If you find more, make the adjustment. There are 2 more that are exactly 26 but they are not counted because the directions say "less than 26."

So set up your proportion

x/2000 = 2/40                     Multiply both sides by 2000

x = 2/40 * 2000

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A

I don't know where 5 comes from. But it is not correct.

B

B should be the correct answer.

C

Exactly 100 pieces should be defective. That is the theoretical result. C is incorrect.

D

D is not correct. The sample size would not be 40. It would have to be 2000 for D to be correct. So D is wrong.

E

We have enough data to get an answer. E is incorrect.

Problem 2

The think you must NOT do is count 1 as being prime. The prime numbers are 2 3 5 7 between 1 and 8. They break down as follows.

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  • 2                         3
  • 3                         4
  • 5                         2
  • 7                         3

The total number of primes = 12

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The experimental probability of tossing a prime is 12/20 * 100% = 60%

The non primes are 2 3 5 7 which is 4 out of 8

4/8 * 100 = 50%

The experimental value is 10% more than the theoretical value.

Discussion

Note: the problem may be one. This all depends on what you have been told about 1. I am using the exact wording of prime here. 1 is not a prime. It is also not a composite. So it has to be counted as part of the non primes.


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