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3 years ago

Which of the following is not equal to the other values? cos31.7° cos211.7° cos328.3° cos(-391.7°)

Mathematics

1 answer:

luda_lava [24]3 years ago

6 0

Answer:

* cos 211.7 not equal the other values

Step-by-step explanation:

* Lets revise the angles in the four quadrant

- If the angle in the first quadrant is Ф, then the equivalent angles to

it in the other three quadrant are

# 180° - Ф ⇒ 2nd quadrant (sin only +ve)

# 180° + Ф ⇒ 3rd quadrant (tan only +ve)

# 360° - Ф ⇒ 4th quadrant (cos only +ve)

# -Ф ⇒ 4th quadrant (cos only +ve)

# -180 + Ф ⇒ 3rd quadrant (tan only +ve)

# -180 - Ф ⇒ 2nd quadrant (sin only +ve)

# -360 + Ф ⇒ 1st quadrant (all are +ve)

* Lets solve the problem

∵ Ф = 31.7°

∵ cos 31.7 = +ve value

∵ 180° + Ф° = 180° + 31.7° = 211.7°

∵ cos (180° + Ф°) = - cos Ф° ⇒ cos (180° + Ф°) in the 3rd quadrant is

same value as cos Ф but with -ve sign

∴ cos 211.7° = - cos 31.7°

∴ cos 31.7° ≠ cos 211.7°

∵ 360° - Ф° = 360° - 31.7° = 328.3°

∵ cos (360° - Ф°) = cos Ф° ⇒ cos (360° - Ф°) in the 4th quadrant has the

same value and sign with cos Ф°

∴ cos 328.3° = cos 31.7°

∴ cos 31.7° = cos 328.3°

∵ -391.7° + 360° = -31.7° ⇒ more then clockwise turn by 31.7°

∵ cos (-Ф°) = cos Ф° ⇒ cos (-Ф°) in the 4th quadrant has the same value

and sign with cos Ф°

∴ cos (-31.7°) = cos 31.7°

∴ cos 31.7° = cos (-390.7°)

* cos 211.7 not equal the other values

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(b) 0.20

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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