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Nadusha1986 [10]
1 year ago
15

Determine whether the given triangle has no solution, one solution or two solutions. Then solve the triangle. Round measures of

sides to the nearest tenth and measures of angles to the nearest degree
A = 119°, a=7, b=4




Question 7 options:

one solution; c ≈ 7; B = 30°; C = 119°

no solution

one solution; c ≈ 4; B = 31°; C = 30°

one solution; c ≈ 4; B = 30°; C = 31°

Mathematics
1 answer:
yuradex [85]1 year ago
7 0

Answer:

The triangle has one solution. The remaining side c ≈ 4 and remaining angles B = 30°; C = 31°.

Option D is correct.

Step-by-step explanation:

if angle A is obtuse and if a > b then the triangle has one solution

We are given ∠ = 119° which is obtuse and side a= 7 and side b - 4 i.e 7>4 so, the triangle has one solution.

Finding remaining side c and ∠B and ∠C

Using Law of sines to find ∠B

a/sin A = b/sin B

7/sin 119° = 4/sin B

7 * sin B = 4 * sin 119

7*sin B = 4(0.874)

sin B = 3.496/7

B = sin^-1(0.4994)

B = 29.96 = 30°

We know that sum of angles of triangle = 180°

So, 180° = 119° + 30° +∠C

180° = 149° + ∠C

=> ∠C = 180° - 149°

∠C = 31°

Now finding c

b/sin B = c /sin C

4/Sin 30 = c/sin 31

4* sin 31 = c*sin 30

4*0.515 = c* 0.5

=> c =  4*0.515/0.5

c = 4.12 ≈ 4

So, Option D one solution; c ≈ 4; B = 30°; C = 31° is correct.

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Step-by-step explanation:

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4 0
1 year ago
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Hotel 3 provides a 6 hour booking for 120 delegates in one room (including tea and lunch).how much would this cost?
satela [25.4K]

Answer:

  $3840

Step-by-step explanation:

The total cost is ...

  (6 hr)(cost per hr) +(120 persons)(food cost per person)

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5 0
2 years ago
The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. Find the probability
yawa3891 [41]

<u>Complete Question</u>

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Find the probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

Answer:

0.9875

Step-by-step explanation:

Total Number of Guests which forms the Sample Space, n(S)=80

Let the Event (a friend of the bride) =A

Let the Event (a friend of the groom) =B

n(A) =59

n(B)=50

Friends of both bride and groom, n(A \cap B)=30

Therefore:

n(A \cup B)=n(A)+n(B)-n(A \cap B)\\n(A \cup B)=59+50-30\\n(A \cup B)=79

The number of Guests who was a friend of the bride OR of the groom = 79

Therefore:

The probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

P(A \cup B) =\dfrac{n(A \cup B) }{n(S)} \\\\=\dfrac{79 }{80}\\\\=0.9875

4 0
1 year ago
marty has 80 to spend at a sporting goods store. he will spend 56 on a shirt, and then buy some darts. each box of darts cost 6.
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Next time, Shay, please include the instructions. I'm assuming that our job here is to determine the largest # of boxes of darts that can be purchased.


Shirt cost + x(box of darts cost) = Initial amount

$56 + x($6/box) = $80


Then 56 + 6x = 80, or 6x= 24, or x = 4. He can buy 4 boxes of darts.


8 0
2 years ago
Keith tabulated the following values for time spent napping in minutes of six of his friends: 23, 35, 17, 30, 20, and 19. The st
il63 [147K]

Answer:

The  t-statistic is t  =  0.6956

Step-by-step explanation:

From the question we are told that

     The population mean is  \mu =  22 \ minutes

       The standard deviation is  s =  7.043

       The  given data is  23, 35, 17, 30, 20, and 19.

Generally the sample mean is mathematically evaluated as

      \=  x  =  \frac{23+  35+17+ 30+ 20+ 19 }{6}

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The t-statistic is mathematically evaluated as

       t  =  \frac{ \= x  - \mu }{\frac{s}{ \sqrt{n} } }

=>   t  =  \frac{ 24   - 22 }{\frac{ 7.043}{ \sqrt{6} } }

=>   t  =  0.6956

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