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4 years ago

In △ABC, a=31, b=22, and c=18. Identify m∠A rounded to the nearest degree. HELP ME PLEASE!!

Mathematics

1 answer:

4vir4ik [10]4 years ago

3 0

Answer:

The measure of angle A is $101\°$

Step-by-step explanation:

we know that

Applying the law of cosines

$a^{2} =b^{2}+c^{2}-2(b)(c)cos(A)$

substitute the values and solve for cos(A)

$31^{2} =22^{2}+18^{2}-2(22)(18)cos(A)$

$cos(A)=[22^{2}+18^{2}-31^{2}]/(2(22)(18))\\ \\cos(A)=-0.193182\\ \\A=arccos(-0.193182)=101\°$

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The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

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(a) What is the expected number of non-defective units?

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$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

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4 years ago