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Kaylis [27]
2 years ago
14

Getaway Travel Agency surveyed a random sample of 454545 of their clients about their vacation plans. Of the clients surveyed, 2

12121 expected that they would go on 333 vacations in the next year. There are 516516516 Getaway Travel Agency clients.
Based on the data, what is the most reasonable estimate for the number of Getaway Travel Agency clients who expect to go on 333 vacations in the next year?
Mathematics
1 answer:
Sergio [31]2 years ago
5 0

Answer:

The most reasonable estimate for the number of Getaway Travel Agency clients who expect to go on 3 vacations in the next year is 241.

Step-by-step explanation:

The best estimate for the proportion of clients who expect to go on 3 vacations is the same proportion of the sample.

Sample:

45 clients.

Of those, 21 expected to go on 3 vacations in the next year.

21/45 = 0.4667

Out of the 516 clients:

0.4667*516 = 240.8

Rounding up

The most reasonable estimate for the number of Getaway Travel Agency clients who expect to go on 3 vacations in the next year is 241.

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A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
2 years ago
Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to
RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

5 0
2 years ago
A broker has a four-year contract to manage a rental unit for an investor. He is to receive 50% of the first month's rent and 5%
Alex777 [14]

Answer:

Step-by-step explanation:

600 X .50 = 300.00 1st month

600 X .05 = 30.00 47 months

30 X 47 = 1,410

1,410 + 300 = 1,710 (final answer)

8 0
2 years ago
Read 2 more answers
Molly bought 12.5 yards of fabric for 4.50 a yard to make dog beds. She uses 2.5 yards of fabric for each dog bed. She sells eac
podryga [215]
I think u just have to multiply
4 0
2 years ago
The area of the trapezoid is 24 and the bases are 3 and 5. find the height. 3 3.2 6
lorasvet [3.4K]
<span>1. To solve this problem you must apply the formula for calcúlate the area of a trapezoid, which is shown below: 
 
 A=(B+b)h/2
 
 A is the area of the trapezoid (A=24).
 B is the larger base </span>of the trapezoid <span>(B=5).
 b is the smaller base </span>of the trapezoid<span> (b=5).
 h is the height </span><span>of the trapezoid.
 
 2. You must clear "h" from the formula:
</span><span> 
 A=(B+b)h/2
</span> 2A=<span>(B+b)h
</span> h=2A/<span>(B+b)
</span><span> 

  3. When you susbtitute the values into </span> h=2A/(B+b), you obtain:
 
 h=2A/<span>(B+b)
</span><span> h=2(24)/(5+3)
 h=48/8
 h=6
</span>
6 0
2 years ago
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