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BabaBlast [244]

3 years ago

8

6)a garden was square, but last year she made it makes or shortening one side 5 meters and the other 8 meters. If the area of th

e smaller garden is 180 m², then what was the area of her square garden

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

let dimention of original garden be x

then (x-8)(x-5) = 180

x^2 -13x -140 = 0

(x-20)(x+7)=0

x= 20 or x = -7

x = 20

original area = (20)^2

= 400 m^2

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4 years ago

J.k. Rowling and R.L. Stone are both ready hunger games . J.k. Reads 35 pages every 2 hours and R.L. Reads 45 pages every 3 hour

san4es73 [151]

Answer:

The rate of change for each reader is :

For J K Rowling : 17 and a half pages per hour

For R L Stone : 15 pages per hour

Step-by-step explanation:

Here the reading rate of each reader is given as:

J.K. Reads 35 pages every 2 hours.

So, the number of pages read by J K in 2 hours = 35 pages

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4 years ago

Does anyone know answer to this ?

yKpoI14uk [10]

Answer:

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8 0

3 years ago

.<br> Which line is parallel to the line that passes through the points of (2,-5) and (-4,1)

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Answer:

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2 years ago

(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the

goblinko [34]

Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

$98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\ \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\ \\ =10.0023\pm2.33(0.000089) \\ \\ =10.0023\pm0.00021 \\ \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\ \\ =(10.0021, \ 10.0025)$

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.

Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

$\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\ \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\ \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\ \\ \Rightarrow n=4.66^2=21.7156\approx22$

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>

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4 years ago