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Otrada [13]
3 years ago
12

The difference between the prediction interval and the confidence interval is that A. the prediction interval is narrower than t

he confidence interval. B. the prediction interval provides an interval estimation for the expected value of y while the confidence interval does it for a particular value of y. C. the prediction interval provides an interval estimation for a particular value of y while the confidence interval does it for the expected value of y. D. the confidence interval is wider than the prediction interval.
Mathematics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

<em>The prediction interval provides an interval estimation for a particular value of y while the confidence interval does it for the expected value of y. </em>

Step-by-step explanation:

<em>A</em><em>. the prediction interval is narrower than the confidence interval.</em>

the prediction interval is always wider than the confidence interval.

<em>B</em><em>. the prediction interval provides an interval estimation for the expected value of y while the confidence interval does it for a particular value of y.</em>

False

<em>C</em><em>. the prediction interval provides an interval estimation for a particular value of y while the confidence interval does it for the expected value of y. </em>

<em>True</em>

<em>D.</em><em> the confidence interval is wider than the prediction interval.</em>

the prediction interval is  wider

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Answer:

See Explanation Below

Step-by-step explanation:

Your question isn't clear; However, I'll solve in two ways

1. Diameter = 14 cm

2. Radius = 14 cm

<em></em>

<em>When Diameter = 14 cm</em>

Given

Area = \pi r^2

Diameter = 14cm

Required

Calculate the area of the circle

First, the radius has to be solved;

Radius = \frac{1}{2} Diameter

Radius = \frac{1}{2} * 14cm

Radius = 7cm

Using the given formula;

The area is as follows

Area = \pi r^2

Area = 3.14 * 7^2

Area = 3.14 * 49

Area = 153.86cm^2

<em>When Radius = 14 cm</em>

Given

Area = \pi r^2

Radius = 14cm

Required

Calculate the area of the circle

Using the given formula;

The area is as follows

Area = \pi r^2

Area = 3.14 * 14^2

Area = 3.14 * 196

Area = 615.44cm^2

7 0
3 years ago
PLS HELP !!! I NEED HELP PLS ILL APPRECIATE IT SM!
bazaltina [42]

Answer:

\frac{3}{4}

Step-by-step explanation:

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5 0
2 years ago
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Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
Ipatiy [6.2K]

Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

(C) y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}

(D) y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R},

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

(D) Using C, let be y=me^{3t} we obtain that it must satisfies 3^2m-4m=1\Rightarrow m=1/5 and then the general solution is y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R}, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).  

4 0
3 years ago
The schools sport Teams call the bank with the teams I charge $125 for the rental of the whole plus $12 for each meal so the tot
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Answer:

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Step-by-step explanation:

From the given information

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Let the number of people attending the banquet=x

Therefore:

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12x=545-125

12x=420

x=420/12=35

Since x=35, we conclude that 35 People attended the banquet.

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dlinn [17]
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2 years ago
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