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Tamiku [17]
3 years ago
7

On a piece of paper, use a protractor to construct △ABC with m∠A=30° , m∠B=60° , and m∠C=90° .

Mathematics
2 answers:
Naddika [18.5K]3 years ago
6 0

Answer:

C BC>AB

D BC<AC

Step-by-step explanation:

Please Give Me A Good Rating And Press Thank You

ANEK [815]3 years ago
5 0

Answer:

C) BC < AB

D) BC < AC

Ste p-by-step explanation:

We are given the following information in the question:

\angle A = 30^\circ\\\angle B = 60^\circ\\\angle C = 90^\circ

This means that the side opposite to the right angle in the triangle is AB.

Since, the side opposite to the right angle is the hypotenuse and is the longest side of the triangle, thus AB cannot be shorter than any other side of the triangle.

Hence, option C is correct

BC < AB

Also, in the triangle we have,

tan~ 60^\circ = \displaystyle\frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} = \frac{\sqrt3}{1}\\\\\Rightarrow AC > BC

Hence, option D) is correct

BC < AC

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The points have to have the coordinate y=4 to be on the y=4 line

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3 years ago
Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the
Levart [38]

Answer: Choice A.)   88.2 < μ < 93.0

=============================================================

Explanation:

We have this given info:

  • n = 92 = sample size
  • xbar = 90.6 = sample mean
  • sigma = 8.9 = population standard deviation
  • C = 99% = confidence level

Because n > 30 and because we know sigma, this allows us to use the Z distribution (aka standard normal distribution).

At 99% confidence, the z critical value is roughly z = 2.576; use a reference sheet, table, or calculator to determine this.

The lower bound of the confidence interval (L) is roughly

L = xbar - z*sigma/sqrt(n)

L = 90.6 - 2.576*8.9/sqrt(92)

L = 88.209757568781

L = 88.2

The upper bound (U) of this confidence interval is roughly

U = xbar + z*sigma/sqrt(n)

U = 90.6 + 2.576*8.9/sqrt(92)

U = 92.990242431219

U = 93.0

Therefore, the confidence interval in the format (L, U) is approximately (88.2, 93.0)

When converted to L < μ < U format, then we get approximately 88.2 < μ < 93.0 which shows that the final answer is choice A.

We're 99% confident that the population mean mu is somewhere between 88.2 and 93.0

7 0
3 years ago
Lis 10 múltiples of 8
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6 0
3 years ago
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8 0
4 years ago
g Suppose that a die is rolled twice. What are the possible values that thefollowing random variables can take on:(a) the maximu
KengaRu [80]

Answer:

(a) A = {1, 2, 3, 4, 5, 6}

(b) B = {1, 2, 3, 4, 5, 6}

(c) C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(d) D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

Step-by-step explanation:

Assume each roll can result in the numbers 1, 2, 3, 4, 5, or 6.

(a) If both rolls result in a 1, the maximum value is 1. If either roll results in a 6, the maximum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

A = {1, 2, 3, 4, 5, 6}

(b) If either roll results in a 1, the minimum value is 1. If both rolls result in a 6, the minimum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

B = {1, 2, 3, 4, 5, 6}

(c) If both rolls result in a 1, the sum is 2. If both rolls results in a 6, the sum is 12; all integers between 2 and 12 are also possible. Therefore, the possible values are:

C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(d) If the first roll results in a 1 and the second results in a 6, the result is -5. On the other hand, if the first roll results in a 6 and the second results in a 1, the result is 5; all integers between -5 and 5 are also possible. Therefore, the possible values are:

D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

7 0
3 years ago
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