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Alexeev081 [22]
3 years ago
15

Imagine two line segments where each represents a slant height of the cone. The segments are on opposite sides of the cone and m

eet at the apex. Find the measurement of the angle formed between the line segments.

Mathematics
2 answers:
sladkih [1.3K]3 years ago
8 0

Imagine that we cut the cone in half with a vertical plane. In this plane we'd see points a, b and c. We need to find the measure of angle bca, and then double that angle, to obtain the answer to this problem.

Note that tan (angle bca) = opp / adj = 2 / 6, or 1/3, or 0.333...

Working backwards to find angle bca, we use the inverse tangent function:

arctan 0.3333.. = approx. 0.322 radians, or 18.435 degrees.

Doubling that gives us the measurement of the angle formed between the line segments: 2(18.435 deg) = 36.9 degrees, approximately.

noname [10]3 years ago
5 0

Imagine that we cut the cone in half with a vertical plane. In this plane we'd see points a, b and c. We need to find the measure of angle bca, and then double that angle, to obtain the answer to this problem.

Note that tan (angle bca) = opp / adj = 2 / 6, or 1/3, or 0.333...

Working backwards to find angle bca, we use the inverse tangent function:

arctan 0.3333.. = approx. 0.322 radians, or 18.435 degrees.

Doubling that gives us the measurement of the angle formed between the line segments: 2(18.435 deg) = 36.9 degrees, approximately.

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Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the
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Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million)     Number of companies

25 up to 35                                                    4

35 up to 45                                                    19

45 up to 55                                                    27

55 up to 65                                                    16

65 up to 75                                                     9

TOTAL                                                            75

Let's find the probabilities first:

P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}

For 35 up to 45

P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}

For 45 up to 55

P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}

For 55 up to 65

P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}

For 65 up to 75

P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                (O-E)^2   \dfrac{(O-E)^2}{E}

             compa                                 cted E (n*p)

             nies (O)  

25-35           4      0.0612    75*0.0612 = 4.59        0.3481       0.0758

35-45           19     0.2218   75*0.2218 = 16.635     5.5932       0.3362

45-55           27     0.3568   75*0.3568 = 26.76     0.0576      0.021

55-65           16      0.2552   75*0.2552 = 19.14      9.8596      0.5151

65-75           9      0.0839     75*0.0839 = 6.2925   7.331         1.1650

                                                                                           \sum \dfrac{(O-E)^2}{E}= 2.0492                                                                                                      

Using the Chi-square formula:

X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square  \ X^2 = 2.0942

Null hypothesis:

H_o: \text{The population of advertising expenses follows a normal distribution}

Alternative hypothesis:  

H_a: \text{The population of advertising expenses does not follows a normal distribution}

Assume that:

\alpha = 0.02

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from X^2 = 11.667

Decision rule: To reject H_o  \  if \  X^2  test statistics is greater than X^2 tabulated.

Conclusion: Since X^2 = 2.0942 is less than critical value 11.667. Then we fail to reject H_o

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