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netineya [11]
3 years ago
13

A 30 gram sample of a substance that's used to sterilize surgical instruments has a k-value of 0.1235. Find the substances half

life in days. Round your answer to the nearest tenth.
Mathematics
1 answer:
jok3333 [9.3K]3 years ago
5 0

Answer : The substances half life in days is, 5.6 days.

Step-by-step explanation :

Half life : It is defined as the amount of time taken by a radioactive material to decay to half of its original value.

All radioactive decays follow first order kinetics.

The relation between the half-life and rate constant is:

k=\frac{0.693}{t_{1/2}}

where,

k = rate constant = 0.1235 per days

t_{1/2} = half-life

Now put all the given values in the above formula, we get:

0.1235\text {days}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=5.6\text{ days}

Thus, the substances half life in days is, 5.6 days.

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The Wind Mountain excavation site in New Mexico is an important archaeological location of the ancient Native American Anasazi c
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Answer:

a) 28

b)

Class Limits  Class Boundaries   Midpoint    f      Rf         Cf

10 - 37            9.5-37.5                   23.5            7     0.096     7      

38 - 65           37.5-65.5                 51.5           25    0.342     32    

66 - 93           65.5-93.5                 79.5          26    0.356     58    

94 -  121        93.5-121.5                   107.5         9     0.123       67    

122 - 149      121.5-149.5                  135.5         5      0.068     72  

150 - 177       149.5-177.5                  163.5        0         0.0      72    

178 - 205       177.5-20.5                  191.5         1       0.014      73  

Total                                                                  73          1.0    

c)

The histogram is in attached file.

Step-by-step explanation:

a)

The width of class interval=range/desired number of classes

The width of class interval=(maximum-minimum)/7

The width of class interval=(200-10)/7

The width of class interval=190/7

The width of class interval=27.14=28 (rounded to next whole number)

Part(b) and part (c) is explained in attached word document.

Download docx
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Find f(a), f(a+h), and<br> 71. f(x) = 7x - 3<br> f(a+h)-f(a)<br> h<br> if h = 0.<br> 72. f(x) = 5x²
Leni [432]

Answer:

71. \ \ \ f(a) \  = \  7a \ - \ 3; \ f(a+h) \  =  \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7

72. \ \ \ f(a) \  = \  5a^{2}; \ f(a+h) \  =  \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

                                              \displaystyle\frac{f(x+h) \ - \ f(x)}{h},

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

                                             m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}.

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as <em>h</em> approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

              71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7

               72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h

4 0
2 years ago
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