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Furkat [3]
3 years ago
11

The half-life of strontium -90 is approximately 29 Year’s . How much of a 150 g sample of strontium-90 will remain after 116 yea

rs
Mathematics
1 answer:
marin [14]3 years ago
3 0
The half-life is 29 years.
116 : 29 = 4
So we will have 4 periods of half-life:
( 1/2 )^4 = 1/16
150 g * 1/16 = 9.375 g
Answer:
9.375 g of strontium -90 will remain after 116 years.
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BartSMP [9]

Answer:

Length\ of\ the\ long\ piece=40\ meters\\\\Length\ of\ the\ short\ piece=17\ meters

Step-by-step explanation:

Let be "s" the length in meters of the short piece and "l" the lenght in meters of the long piece.

Set up a system of equations:

\left \{ {{s+l=57} \atop {s=l-23}} \right.

Apply the Substitution Method to solve the system. Substitute the second equation into the first equation and solve for "l":

(l-23)+l=57\\\\2l=80\\\\l=\frac{80}{2}\\\\l=40

Susbstitute the value of "l" into the second equation in order to find the value of "s". This is:

s=40-23\\\\s=17

6 0
3 years ago
What are the missing numbers?
n200080 [17]
2^2. 2 and 2. Because 2 and 2
8 0
3 years ago
Please answer right away
irina1246 [14]

Answer:

$29000 with a margin of error of $5000

Step-by-step explanation:

We have that the midpoint between the given values ​​is

(X1+X2) / 2 = ($34000+$24000)/2 = $29000

We have that the midpoint between the given values ​​would be

(X2-X1)/2=($34000-$24000)/2=$10000/2=$5000

So I can write that approach as $29000 with a margin of error of $5000

Done

5 0
3 years ago
In a sample of 679 new websites registered on the Internet, 42 were anonymous (i.e., they shielded their name and contact inform
Semmy [17]

Answer:

95% Confidence interval:  (0.0429,0.0791)      

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 679

Number of anonymous websites, x = 42

\hat{p} = \dfrac{x}{n} = \dfrac{42}{679} = 0.0618

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96

Putting the values, we get:

0.0618\pm 1.96(\sqrt{\dfrac{0.0618(1-0.0618)}{679}}) = 0.0618\pm 0.0181\\\\=(0.0429,0.0791)

is the required confidence interval for proportion of all new websites that were anonymous.

8 0
3 years ago
What is the percent of change to 125 centimeteres 87.5 centimeters?
MrRa [10]
I hope this helps you

125-87,5=37,5

125 37,5

100 ?

?.125=100.37,5

?=3750/125

?=30


30%
5 0
3 years ago
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