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Alekssandra [29.7K]
3 years ago
15

At a dock on the east coast, low tide occurs at 3 pm wi th a water depth of 5 ft. The depth at high tide is 83 ft. High tide occ

urs every 6 hours. Explain how to find the sinusoidal function that models the depth in terms of time, x.
Mathematics
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

The sinusoidal function that model the depth in terms of time, 'x', is presented here as follows;

f(x) = 39·sin((π/3)·x - π) + 5

Step-by-step explanation:

A sinusoidal function is given by the general function as follows;

y = A·sin(B·x + C) + D

Where;

A = The amplitude = (y_{max} - y_{min})/2 = (The high tide - The low tide)/2 = (83 ft. - 5 ft.)/2 = 39 ft.

The period, T = 2·π/b = 6 hours

∴ B = 2·π/T = 2·π/6 = π/3

D = The vertical shift = The low tide = 5 ft.

The horizontal phase shift, 'C', is given as follows;

3 hrs = -C/B = -C/(π/(3 hr))

C = 3 hrs × -(π/(3 hr)) = -π

∴ C = -π

y = 39·sin((π/3)·x - π) + 5

The sinusoidal function that model the depth in terms of time, 'x', is therefore given as follows;

f(x) = 39·sin((π/3)·x - π) + 5

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Answer:

540m

Step-by-step explanation:

From the question,we are told

It takes 30 minutes or less to walk from home to school and back home again

Time walking to school = 30 or less minutes

Time walking from school = 30 less minutes

Speed walking to school = 30m/min

Speed walking from school = 45m/min

The question asked us to find the Distance between home and school

The formula for Distance covered = Speed × Time

Total distance = Total Speed(m/min) × Total time(minutes)

Let the distance between home and school be represented by x

Time to walk from the house to school = x/ 45m/min + x/ 30m/min

= 30 × x + 45 × x/1350

=( 75x / 1350)

Time to walk from school or to school = 30 minutes of less

= 75x/1350 = 30

75x = 1350 × 30

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Answer:

1.   20 cm^2

2.  40 cm^2

3.   14 cm^2

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Step-by-step explanation:

Area of triangle= 1/2(base)*(height)

Area of rectangle= (base)*(height)

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Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

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