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DIA [1.3K]
3 years ago
14

How many solutions are possible for a triangle with A = 113° , a = 15, and b = 8

Mathematics
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

One solution.

Step-by-step explanation:

To determine the number of possible solutions for a triangle with A = 113° , a = 15, and b = 8, we're going to use the law of sines which states that: "<em>When we divide side a by the sine of angle A  it is equal to side b divided by the sine of angle B,  and also equal to side c divided by the sine of angle C</em>".

Using the law of sines we have:

\frac{sin(A)}{a} = \frac{sin(B)}{b}

\frac{sin(113)}{15} = \frac{sin(B)}{8}

Solving for B, we have:

sin(B)=0.4909

∠B = 29.4°

Therefore, the measure of the third angle is: ∠C = 37.6°

There is another angle whose sine is 0.4909 which is 180° - 29.4° = 150.6 degrees. Given that the sum of all three angles of any triangle must be equal to 180 deg, we can't have a triangle with angle B=113° and C=150.6°, because B+C>180.

Therefore, there is one triangle that satisfies the conditions.

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4. A small high school holds its graduation ceremony in the gym. Because of seating constraints, students are limited to a maxim
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Answer:

(a) The mean and standard deviation of <em>X</em> is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of <em>T</em> are 390 and 180 respectively.

(c) The distribution of <em>T</em> is <em>N</em> (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.

Step-by-step explanation:

(a)

The random variable <em>X</em> is defined as the number of tickets requested by a randomly selected graduating student.

The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:

X      P (X)

0      0.05

1       0.15

2      0.25

3      0.25

4      0.30

The formula to compute the mean is:

\mu=\sum x\cdot P(X)

Compute the mean number of tickets requested by a student as follows:

\mu=\sum x\cdot P(X)\\=(0\times 0.05)+(1\times 0.15)+(2\times 0.25)+(3\times 0.25)+(4\times 0.30)\\=2.6

The formula of standard deviation of the number of tickets requested by a student as follows:

\sigma=\sqrt{E(X^{2})-\mu^{2}}

Compute the standard deviation as follows:

\sigma=\sqrt{E(X^{2})-\mu^{2}}\\=\sqrt{[(0^{2}\times 0.05)+(1^{2}\times 0.15)+(2^{2}\times 0.25)+(3^{2}\times 0.25)+(4^{2}\times 0.30)]-(2.6)^{2}}\\=\sqrt{1.44}\\=1.2

Thus, the mean and standard deviation of <em>X</em> is 2.6 and 1.2 respectively.

(b)

The random variable <em>T</em> is defined as the total number of tickets requested by the 150 students graduating this year.

That is, <em>T</em> = 150 <em>X</em>

Compute the mean of <em>T</em> as follows:

\mu=E(T)\\=E(150\cdot X)\\=150\times E(X)\\=150\times 2.6\\=390

Compute the standard deviation of <em>T</em> as follows:

\sigma=SD(T)\\=SD(150\cdot X)\\=\sqrt{V(150\cdot X)}\\=\sqrt{150^{2}}\times SD(X)\\=150\times 1.2\\=180

Thus, the mean and standard deviation of <em>T</em> are 390 and 180 respectively.

(c)

The maximum number of seats at the gym is, 500.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Here <em>T</em> = total number of seats requested.

Then, the mean of the distribution of the sum of values of X is given by,  

\mu_{T}=n\times \mu_{X}=390  

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{T}=n\times \sigma_{X}=180

So, the distribution of <em>T</em> is N (390, 180²).

Compute the probability that all students’ requests can be accommodated, i.e. less than 500 seats were requested as follows:

P(T

Thus, the probability that all students’ requests can be accommodated is 0.7291.

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=> AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} }  }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2}  }=\sqrt{5^{2}+8^{2}  }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2}  }=\sqrt{65}

using Heron theorem, we have:

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with S is the area of the triangle

       p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65}  }{2}

Step-by-step explanation:

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LMF AO, I would love to help you but there is no answer choices....LOL

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