More Algebra 2 problems.

__August 2017, Part I__

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.

*10. lridium-192 is an isotope of iridium and has a half-life of 73.83 days. If a laboratory experiment begins with 100
grams of Iridium-192, the number of grams, A, of Iridium-192 present after t days would be A = 100(1/2) ^{(t/73.83)}.
Which equation approximates the amount of Iridium-192 present after t days?
1) A = 100(73.83/2)^{t}
2) A = 100(1/147.66)^{t}
3) A = 100(0.990656)^{t}
4) A = 100(0.116381)^{t}
*

**Answer: 3) A = 100(0.990656) ^{t}**

100(1/2)

^{(t/73.83)}= 100((1/2)

^{(1/73.83)})

^{(t)}

(1/2)

^{(1/73.83)}= 0.99065551184

So 100(1/2)

^{(t/73.83)}= 100(0.990656)

^{(t)}

*11. The distribution of the diameters of ball bearings made under a given manufacturing process is normally
distributed with a mean of 4 cm and a standard deviation of 0.2 cm. What proportion of the ball bearings will
have a diameter less than 3.7 cm?
1) 0.0668
2) 0.4332
3) 0.8664
4) 0.9500
*

**Answer: 1) 0.0668**

Enter *normCdf(0,3.7,4,0.2)* into your calculator.

You'll find "normCdf" using 2nd VARS, which gives you the DISTRIB menu, where it is option 2.

The 0 is your lower boundary, and 3.7 is your upper boundary. You want everything less than 3.7.

The 4 is your mean, and the 0.2 is your standard deviation.

You will get an answer of 0.066807.

*12. A polynomial equation of degree three, p(x), is used to model the volume of a rectangular box. The graph of p(x)
has x intercepts at −2, 10, and 14. Which statements regarding p(x) could be true?
A. The equation of p(x) = (x − 2)(x + 10)(x + 14).
B. The equation of p(x) = −(x + 2)(x − 10)(x − 14).
C. The maximum volume occurs when x = 10.
D. The maximum volume of the box is approximately 56
1) A and C
2) A and D
3) B and C
4) B and D
*

**Answer:4) B and D**

The zeroes are -2, 10 and 14, so statement A is incorrect but B could be true.

Since 10 is a zero, it does not make sense that it would also be a maximum value. The line will be either increasing or decreasing as it passes through x = 10. In either case, x = 10 is not a maximum point. This eliminates statement C, so only Choice 4 is left.

Putting p(x) = −(x + 2)(x − 10)(x − 14) into the calculator, you will see that the maximum is 56, so statement 4 could be true if statement 2 is true.

Comments and questions welcome.

More Algebra 2 problems.

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