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user100 [1]
2 years ago
8

A research organization wanted to estimate the average number of hours a college student sleeps per night during the school year

. After randomly sampling 150 college​ students, the research organization determined the following​ 95% confidence​ interval: (7.1,7.5)​ hours/night. What is the value of the average number of hours slept per night during the school year for these 150 college​ students?
Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
4 0

Answer:

Step-by-step explanation:

Given given that

  • Lower confidence interval = 7.1
  • Upper confidence interval = 7.5
  • As such, We are 95% confident that it's somewhere between 7.1 and 7.5 hours per night

Average value = LCI + UCI /2 = 7.1 + 7.5 / 2

= 7.3hours per night, so we are 95% confident that it's somewhere between 7.1 and 7.5 hours per night which is eventually 7.3hrs per night.

You might be interested in
A theatre has the capacity to seat people across two levels, the Circle and
andriy [413]

Answer: 76.19\%

Step-by-step explanation:

<h3> The complete exercise is: " A theatre has the capacity to seat people across two levels, the Circle, and the stalls. The ratio of the number of seats in the circle to a number of seats in the stalls is 2:5. Last Friday, the audience occupied all the 528 seats in the circle and \frac{2}{3} of the seats in the stalls. What is the percentage of occupancy of the theatre last Friday?"</h3>

Let be "s" the total number of seats in the Stalls.

The problem says that the ratio of the number of seats in the Circle to the number of seats in the Stalls is 2:5.

Since the number of seats that were occupied last Friday was 528 seats, we can set up the following proportion:

\frac{2}{5}=\frac{528}{s}

Solving for "s", we get:

s*\frac{2}{5}=528\\\\s=528*\frac{5}{2}\\\\s=1,320

So the sum of the number of seats in the Circle and the number of seats in the Stalls, is:

Total=1,320\ seats+528\ seats=1,848\ seats

 We know that \frac{2}{3} of the seats in the Stalls were occupied. Then, the number of seat in the Stalls that were occupied is:

(1,320)(\frac{2}{3})=880

Therefore, the total number of seats that were occupied las Friday is:

Total\ occupied=880\ seats+528\ seats=1,408\ seats

Knowing this, we can set up the following proportion, where "p" is the the percentage of occupancy of the theatre last Friday:

\frac{100}{1,848}=\frac{p}{1,408}

Solving for "p", we get:

(1,408)(\frac{100}{1,848})=p\\\\p=76.19\%

8 0
1 year ago
Mrs.Helio has 2 3/4 acres of farmland. she will plant corn on 1/4 of this land, potatoes on 1/12 of the land, wheat on 5/8 of th
Kazeer [188]
2 3/4 acres is the same as 11/4 acres
To find how much of the land is beans, find how much is already taken up, and the rest is beans.
Farmland-Corn-Potatoes-Wheat=Beans
1-(1/4)-(1/12)-(5/8)=1/24

Corn: (1/4)x(11/4)=11/16 acres
Potatoes: (1/12)x(11/4)=11/48 acres
Wheat: (5/8)x(11/4)=55/32 acres or 1 23/32 acres
<span>Beans: (1/24)x(11/4)=11/96 acres</span>
4 0
2 years ago
Korey starts a small carwash business to save up some cash. He decides to offer two different price packages to his clients. Pac
rosijanka [135]

Answer:

Package A

Step-by-step explanation:

Hello there!

The reason that package A is a better deal is beacause it goes down by a fixed rate, not by a percentage. If you get 15 carwashes, you will pay nothing for the 15th, but even if you washed your car 100 times with Package B, you would still pay something!

Have a great day!

5 0
1 year ago
Mark and Sofia walk together down a long, straight road. They walk without stopping for 4 miles. At this point Sofia says their
Marina CMI [18]

Answer:

Sofia is correct.

Step-by-step explanation:

Mark's statement depends on the starting point. He can be correct if they started at the 0 mile, but in this case we don't know where they started. They could had started at the 12 mile and their current position after the walking would be 16 miles.

On the other hand, Sofia's statement doesn't depend on where they started. She refers to how much they walked, not to where they are after the walking. Since they stopped after 4 non-stopping miles, their displacement was exactly 4 miles. So Sofia is correct.

5 0
2 years ago
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
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