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3 years ago

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A plane at an altitude of 7000 ft is flying in the direction of an island . If an angle of depression is 21 ° from the plane to

the island , what is the horizontal distance until the plan flies over the island ?

1 answer:

Nonamiya [84]3 years ago

3 0

The horizontal distance until the plan flies over the island is 2687.05 feet approximately.

<u>Solution:</u>

Given that, A plane at an altitude of 7000 ft is flying in the direction of an island

An angle of depression is 21 degree from the plane to the island

We have to find what is the horizontal distance until the plan flies over the island

The diagram is attached below

Assume as shown in the diagram , now we can use the right angle triangle property

$\tan (21)=\frac{\text {distance between plane and point above island}}{\text {height of plane}}$

$\tan (21)=\frac{\text {distance between plane and point above island}}{7000}$

$\text { Distance between plane and point above island }=7000 \times \tan (21)$

$=7000 \times 0.38386=2687.0482$

Hence, the distance between plane and point above island is 2687.05 feet approximately.

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A jacket is advertised to be 85% off the original price. If the sale price is $102 what is the original price ?

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The answer would prob around $188.70 because .85x102=86.7 so add that to 102 and u get 188.70 but I'm prob wrong and I have a test on this tomorrow

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3 years ago

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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3 years ago

Solve the equations 11(x+10)=132

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11(x+10) = 132

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11x + 110 = 132
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Answer:

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3 years ago

prove that sin theta cos theta = cot theta is not a trigonometric identity by producing a counterexample

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Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Begin with the right hand side:

R.H.S = cot θ = $\frac{cos \ \theta}{sin \ \theta}$

L.H.S = sin θ cos θ

so, sin θ cos θ ≠ $\frac{cos \ \theta}{sin \ \theta}$

So, the equation is not a trigonometric identity.

=========================================================

<u>Anther solution:</u>

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

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3 years ago