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katrin [286]
3 years ago
10

five sides of a decagon each measure 3 in , the perimeter of the decagon is 40 in, what is the link of each remaining side, if e

ach remaining side has the same length.
Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0
The remaining lengths are equal to 5 inches
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So I have another trigonometry question PLZ HELP ME I'm clueless at this rate I have like a little idea but I'm lost, SO PLZZZ!
Sergio039 [100]

Answer:

1) The windsurfer is approximately 580 ft from each lifeguard stand.

2) The distance between the Earth and Mercury is approximately 61 million miles.

Step-by-step explanation:

The image other two situations described is presented in the attached image to this answer.

1) From the attached image, the windsurfer forms a right angled triangle with each of the lifeguard stand and the midpoint between the two lifeguard stands.

Hence, the angle at the top of the triangle is half of 30°, 15° and the distance from the midpoint of the lifeguard stands to the lifeguard stands is 300/2 = 150 ft.

Let the required distance of the windsurfer from each of the lifeguard stands be x.

Using trigonometric relations,

Sin 15° = (150/x)

x = 150 ÷ sin 15° = 150 ÷ 0.2588

x = 579.6 ft = 580 ft to the nearest whole number

2) From the image, the Sun, Earth and Mercury form a triangle.

Let the possible distance between the Earth and Mercury be y.

Using cosine rule,

y² = 36² + 93² - (2×36×93×cos 22°)

y² = 3,736.5769098208

y = √3,736.5769098208 = 61.13 million miles = 61 million miles to the nearest million miles.

Hope this Helps!!!

5 0
2 years ago
I WILL MARK YOU AS BRAINLIEST IF YOU ARE CORRECT. Look at the image below.
Masteriza [31]

Answer:

x=8.06

Step-by-step explanation:

find you half base and use PYTHAGORAS THEOREM to find the opposite side (x)

6 0
3 years ago
Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t
STatiana [176]

Answer:

<em>Part a) Probability that a moose in that age group is killed by a wolf</em>

  • P  (0.5 year) = 0.361
  • P (1 - 5) = 0.159
  • P (6 - 10) = 0.267
  • P (11 - 15) = 0.203
  • P (16 - 20) = 0.010

<em>Part b)</em>

  • <em>Expected age of a moose killed by a wold</em>

         μ = 5.61 years

  • <em>Stantard deviation of the ages</em>

       σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

<u>Age of Moose in years          Number killed by woolves</u>

Calf (0.5)                                                 107

1 - 5                                                           47

6 - 10                                                         79

11 - 15                                                        60

16 - 20                                                        3

You can  now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) <u>First question: </u>a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P  (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) <u>Second question:</u> b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class       midpoint

   0.5               0.5

   1 - 5           (1 + 5) / 2 = 3

   6 - 10         (6 + 10) / 2 = 8

   11 - 15         (11 + 15) / 2 = 13

   16 - 20       (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s =  (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8  - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

7 0
2 years ago
Jim earns $1,600 per month after taxes. He is working on his budget and has the first three categories finished.
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allochka39001 [22]

Step-by-step explanation:

BAC, BCA, CBA, ACG, ABE, EGC, GEB, EDG, EDA are all vertical of this figure.

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3 years ago
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