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2 years ago

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In pensacola in june, high tide was at noon. the water level at high tide was 12 feet and 2 feet at low tide. assuming the next

high tide is exactly 12 hours later and that the height of the water can be modeled by a cosine curve, find an equation for water level in june for pensacola as a function of time (t). f(t) = 12 cospi over 2t 5 f(t) = 5 cospi over 2t 12 f(t) = 5 cospi over 6t 7 f(t) = 7 cospi over 6t 12

1 answer:

castortr0y [4]2 years ago

3 0

An equation for water level in june for pensacola as a function of time (t) is f(t) = 5 cos pi/6 t + 7.

Which equation of cos show period amplitude ?

The equation given below show aplitude and period

$y = A cos bx + c$

where A = amplitude,

b = 2 pi/Period,

Period = 12 hrs,

c = midline,

x = t and y = f(t)

We have to find the amplitude

<h3>What is the formula for the amplitude?</h3>

$A = 1/2 (Xmax - Xmin)$

$12 - 2 / 2 = 10/2 = 5$

$b = 2 pi / 12 = pi/6$

$c = 1/2 (Xmax + Xmin)$

$12+2/2 = 7$

Therefore, the an equation for water level in june for pensacola as a function of time (t)

$f(t) = 5 cos pi/6 t + 7$

To learn more about the function of time visit:

brainly.com/question/24872445

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3 years ago

What is the quotient of b^3+4b^2-3b+126/b+7

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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

How much this will be ......evaluate -4 (9)(-5)

klemol [59]

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3 years ago