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scoray [572]
2 years ago
11

A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list

ed below. Use a 0.05 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute? 70 79 38 63 44 23 62 61 67 50 61 70 94 87 65
Mathematics
1 answer:
Anit [1.1K]2 years ago
8 0

Answer:

Step-by-step explanation:

Given that a group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below.

Data set is as ollows:

70 79 38 63 44 23 62 61 67 50 61 70 94 87 65

Mean = 62.27Variance =333.35std dev = 18.258std error = 4.714

H0: mu = 60 sec

Ha: mu not equals 60 sec

Mean diff = 2.27

Mean = 62.27Variance =333.35std dev = 18.258std error = 4.714

Test statistic = 2.27/SE =0.4815

p value =0.6376

Since p>0.05, we accept null hypothesis

i.e. there is statistical evidence to say that students are reasonably good at estimating one​ minute

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The value of good wine increases with age. Thus, if you are a wine dealer, you have the problem of deciding whether to sell your
Katyanochek1 [597]

Answer: 64 years

Step-by-step explanation:

Let assume the dealer sold the bottle now for $P, then invested that money at 5% interest. The return would be:

R1 = P(1.05)^t,

This means that after t years, the dealer would have the total amount of:

$P×1.05^t.

If the dealer prefer to wait for t years from now to sell the bottle of wine, then he will get the return of:

R2 = $P(1 + 20).

The value of t which will make both returns equal, will be;

R1 = R2.

P×1.05^t = P(1+20)

P will cancel out

1.05^t = 21

Log both sides

Log1.05^t = Log21

tLog1.05 = Log21

t = Log21/Log1.05

t = 64 years

The best time to sell the wine is therefore 64years from now.

7 0
2 years ago
What is the common difference between the elements of the arithmetic sequence below?
fredd [130]
All numbers in each number add up to 9, i dont know if you understand me, 
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2+2+5=9
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3 0
2 years ago
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The formula for the volume of a sphere is mc012-1.jpg. What is the formula solved for r? mc012-2.jpg mc012-3.jpg mc012-4.jpg mc0
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Cannot see your image, but the formula for the volume of a sphere is
V=(4/3)πr³
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2 years ago
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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/∝

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2 years ago
A company logo is made up of a square and three identical triangles. What is the area of the logo? Enter your answer in the box.
garri49 [273]
Given:
square with sides measuring 7 cm.
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Area of a square = s² = (7cm)² = 49 cm²

Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm² 
Area of the 3 triangles = 14 cm² x 3 = 42 cm²

Total area of the logo = 49 cm² + 42 cm² = 91 cm²
7 0
2 years ago
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