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Inessa [10]
3 years ago
12

An element with a mass of 120 grams decays by 27.5% per minute. To the nearest tenth of a minute, how long will it be until ther

e are 2 grams of the element remaining?
Mathematics
1 answer:
tangare [24]3 years ago
7 0

Answer:

12.7

Step-by-step explanation:

its right

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1. The length of a rectangle is 8.7cm.
Shkiper50 [21]
The width is 9.982 cm and the area is 34.07 and the diagonal is 9.52 I have I answered what you’re asking.
7 0
3 years ago
Ten samples of coal from a northern appalachian source had an average mercury content of 0.242 ppm with a standard deviation of
Dovator [93]
The confidence interval is from 0.2191 to 0.2649, given by

0.242\pm0.0229.

The z-score we need is found by first subtracting the confidence level from 1:
1 - 0.95 = 0.05

Divide that by 2:
0.05/2 = 0.025

Subtract that from 1:
1 - 0.025 = 0.975

Look this up in the middle of a z-table (http://www.z-table.com) and we see that the z-score is 1.96.

Now our confidence interval is given by
\overline{x} \pm z\times \frac{\sigma}{\sqrt{n}}
\\
\\0.242\pm 1.96 \times (\frac{0.037}{\sqrt{10}})
\\
\\0.242 \pm .0229
\\
\\0.242-0.0229=0.2191
\\
\\0.242+0.0229 = 0.2649
5 0
3 years ago
The area of the circle is square
Igoryamba

Answer:

16л, 8л

Step-by-step explanation:

r=4

area=4*4л

=16л

circumference= 2л*4

=8л

4 0
2 years ago
Read 2 more answers
Which of the following number sentences is true?<br> A -9=9<br> B-9=-(9)<br> C 9=(-9)<br> D 9=-(-9)
kondaur [170]

Answer:

A and D are the true number sentences

3 0
3 years ago
) Find the coordinates of the point A' which is the symmetric point
soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point  A(3, 2) with respect to the line 2x + y - 12 = 0.​

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

C( \dfrac{x+3}{2}, \dfrac{y+2}{2})

Also, product of slope will be -1 .( Since, they are parallel )

\dfrac{y-2}{x-3} \times -2  = -1\\\\2y - 4 = x - 3\\\\2y - x = 1

x = 2y - 1

So, C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})

Also, C satisfy given line :

2\times ( \dfrac{2y + 2}{2})  + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}

Also,

x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}

Therefore, the symmetric points isA'(\dfrac{31}{5}, \dfrac{18}{5}) .

7 0
3 years ago
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