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Rina8888 [55]
3 years ago
13

canning operation is designed to fill cans with 14.5 oz of ingredients, on average, with a standard deviation of 0.1 oz. Nine ca

ns are selected at random from the process, and the weights of the nine cans averaged 14.52 oz with a standard deviation of 0.075 oz. The standard error of the sample mean is:
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
3 0

Answer:

SE = 0.025

Step-by-step explanation:

We are given;

Sample mean; x¯ = 14.52

Sample standard deviation; s = 0.075

Sample size; n = 9

Now,formula for standard error of sample mean is;

SE = s/√n

SE = 0.075/√9

SE = 0.075/3

SE = 0.025

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Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)
viktelen [127]
\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}

The first line segment can be parameterized by \mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle with 0\le t\le1. Denote this first segment by C_1. Then

\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt
=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt
=108

The second line segment (C_2) can be described by \mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle, again with 0\le t\le1. Then

\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt
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Finally,

\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3
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The triangles shown below must be congruent.<br> A. True<br> B. False
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Step-by-step explanation:

The snowplow's speed is 40 mph minus the loss from the snow, which is 1.2 mph times the depth of snow in inches.

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The snowplow stops moving when the snow is 33 ⅓ inches deep or more.

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