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Cloud [144]
3 years ago
13

Imagine you are responsible for writing an advice column about mathematics. The following question is submitted to your column:

Dear Math Guru, I am in trouble! I was asked to find the area of a triangle but all I know is that the longest side of the triangle is 13 units, the shortest side is 7 units, the third side is 8 units, and two of the angle measures are 32° and 28°. What do I do? How should you respond? As a group, come up with an explanation that will help the student. You may wish to draw some figures of your own and describe them to the class. Be sure you address the problem for the student and provide accurate and detailed information. Work together to determine the best answer for the student.

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Explanation:

There are two different formulas that are useful with the given information:

  Area = (1/2)ab·sin(C)

  Area = √(s(s-a)(s-b)(s-c)) . . . where s=(a+b+c)/2

It does not matter which sides are designated a, b, and c. Angle C will be opposite side c.

The third angle can be computed based on the fact that the sum of angles in a triangle is 180°. It will be 180° -32° -28° = 120°. The least-to-greatest order of the angles is the same as the least-to-greatest order of the length of the opposite side. So, we might have ...

  • a = 7, A = 28°
  • b = 8, B = 32°
  • c = 13, C = 120°

Utilizing the first formula, the area is ...

  Area = (1/2)(7)(8)sin(120°) = 14√3 ≈ 24.249 . . . square units

Utilizing the second formula, the area is ...

  s = (7+8+13)/2 = 14

  Area = √(14(14-7)(14-8)(14-13)) = 14√3 ≈ 24.249 . . . square units

__

For the answer to be complete, it should be noted that using either of the other two angles will give different results for the area. That is because those angles are not exact values, but are rounded to the nearest degree. Using the first formula with the different angles, we get ...

  • area = (1/2)(8)(13)sin(28°) ≈ 24.413 . . . square units
  • area = (1/2)(7)(13)sin(32°) ≈ 24.111 . . . square units

The first of these answers is a little high because 28° is a little more than the actual value of the angle. Likewise, the second of these answers is a little low because 32° is slightly smaller than the actual angle.

In short, the most accurate information available should be used if the answer is to be the most accurate possible. If the angles are exact, then their values should be used. If the side measures are exact, then their values should be used. In general, it will be easier to make accurate measurements of the side lengths than to make accurate angle measurements.

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let p_1 be the probability that a bearing meets the specification.

So, p_1=0.9

Sample size, n_1=500, is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: \mu_1=n_1p_1=500\times0.9=450

Variance: \sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45

\Rightarrow \sigma_1 =\sqrt{45}=6.71

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, X\geq 440.

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56.

So, the probability that a given shipment is acceptable is

P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062

Hence,  the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by p_2.

p_2=0.94

The total number of shipment, i.e sample size, n_2= 300

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, \mu_2, and variance, \sigma_2^2.

Mean: \mu_2=n_2p_2=300\times0.94=282

Variance: \sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92

\Rightarrow \sigma_2=\sqrt(16.92}=4.11.

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85.

So, the probability that a given shipment is acceptable is

P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977

Hence,  the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, \alpha be the required probability of acceptance of one shipment.

So,

-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}

On solving

\alpha= 0.977896

Again, the probability of acceptance of one shipment, \alpha, depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let p be the probability that one bearing meets the specification. So

-2.01=\frac{439.5-500  p}{\sqrt{500 p(1-p)}}

On solving

p=0.9053

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

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Answer:

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