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daser333 [38]
3 years ago
6

May someone please help??

Mathematics
1 answer:
babymother [125]3 years ago
3 0

Answer:

your inequality signs are correct so i'll skip those

-2.06, .9803

.67, .7486

Step-by-step explanation:

1.)

To do this one you need to standardize x by subtracting the mean and dividng by the standard deviation

(1000-3262)/1100= -2.056 which i will round to -2.06

with this as our z value use the table to find the associated probability

.9803

2.)

Same process

(4000-3262)/1100=.67

which has a probability of .7486

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What is 12% of 68? please show work
tankabanditka [31]
12% of 68 is 8.16. In order to find this answer you must change 12% into a decimal (.12) and then multiply by 68. (68×.12=8.16).
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3 years ago
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algebric A Company Manufactures Hex Bolts And Machine Screws In The Ratio 4:9. How Many Fasteners In All Can Be Made, If 963 Mac
mixas84 [53]

Answer: 1391

Step-by-step explanation:

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grandymaker [24]

this is it :

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4 years ago
If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt
MatroZZZ [7]
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t
\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle

with 0\le t\le1. Then we find the differential:

\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle
\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt

with 0\le t\le1.

Here, the line integral is

\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle
=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt
=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt
=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt
=x_1y_2-x_2y_1

as required.
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4 years ago
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Answer:

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