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finlep [7]
1 year ago
8

There are 1,000 golden delicious and 1,000 red delicious apples in a cooler. In a random sample of 75 of the golden delicious ap

ples, 48 had blemishes. In a random sample of 75 of the red delicious apples, 42 had blemishes. Assume all conditions for inference have been met. Which of the following is closest to the interval estimate of the difference in the numbers of apples with blemishes (golden delicious minus red delicious) at a 98 percent level of confidence?A. (–0.076,0.236)B. (–0.105,0.265)C. (–10.5,26.5)D. (–76,236)E. (−105,265)
Mathematics
1 answer:
Delvig [45]1 year ago
4 0

Answer:

98% confidence interval is (-0.105 , 0.265)

Step-by-step explanation:

n1 = 75          n2 = 75

X1 = 48          X2 = 42

P1 = X1/N1 = 0.64          P2 = X2/N2   = 0.56

98% confidence interval for difference between two proportions is :

P1 - P2  <u>+</u>   Z \sqrt{P1 (1-P1)/n1  + P2 (1-P2)/n1}

= 0.08 <u>+</u> 0.185

= (-0.105 , 0.265)

98% confidence interval is (-0.105 , 0.265)

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If a pack of paper costs 3.75 including tax, and there is a $20 budget, then the equation to find out the answer would be 3.75p is greater or equal to $20. So the answer would be B. 3.75 would be multiplied by the tax and then applied up to $20. 
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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
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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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