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A, the BIOS, is incorrect because its function is to translate communications between the processor and other computer components. The CMOS battery ensures the BIOS settings are not lost when the power is turned off. The BIOS settings and the time are both retained because of the CMOS battery but otherwise are not related to one another. B, the system timer, is incorrect because its function is to synchronize the timing of processes in the computer. It works like a metronome, setting the speed of the computer, and is unrelated to the real time. D, the processor, is also incorrect. The processor is the brain of the computer, and although it might be responsible for incorporating the time information into other tasks, it does not keep track of the time, even while the computer is turned on. 5. B. The BIOS configuration settings are stored on the CMOS chips so that they are not lost when the computer is turned off. The CMOS is attached to a small battery, which allows it to retain information when there is no power. When the computer is turned on, the BIOS retrieves its settings from the CMOS. A, the CMOS uses information stored in the BIOS to set computer configurations, is incorrect because it is the CMOS, not the BIOS, that stores computer configurations. C, the CMOS communicates with computer components based on the information stored in the BIOS, is incorrect because this is the opposite of the actual relationship between the BIOS and CMOS. It is the BIOS that communicates with components, based on information stored in the CMOS. D, the BIOS is a newer version of the CMOS, is incorrect because the functions of the BIOS and CMOS are different, and both are found in all computers, old and new.

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Aside frornthe lirnitations a cclmputer's of numbersysteul, actualarithmetic the manipulations involving thesenunrbers can also resultin roundofT error.To understand how this performs occurs. let's look at how the computer simpleadditionanclsubtraction. Because their familiarity,normalized base-10 numbers to of will be ernployed illuserrorson simpleadditionand subtraction. tratethe effectof roundofT Other numberbases would behave a sirrrilar in fashion. simplify the cliscussion, will ernploya hypothet l we ical decimalcomputer with a 4-digit mantissa a l-digit exponent. and When two floating-pointnumbersare addecl, numbersare first expressed that the so they havethe sane exponents. example, we want to add | .5-57 0.0434 thecomFor if l, + puter would express nurnbers 0. 1557x l0r + 0.00+3-ll x l()r.'fhen the nrantissas the as are addedto give 0. 16004 x t0r . Now. because hypothetical I this computer only carries a 4 - d i g i t m a n t i s s a . t h e e x c e s s n u m b e r o f d i g i t s g e t c h o p p e d o f f a n d t h e r e s u l lt 0s . . l 6 0 0 x i l0 Notice how the last two digits of the secondnumber(41) that were shifiedto the right have essentially beenlost fiom the computation. is Subtraction perfbrmed identically additionexceptthatthe sign of the subtrahend to is rcversed. Forexample. suppose that we are subtracting 26.86frorn 36.41.That is, 0.-164x 102 1 -0.2686 x 102 x 0.095-5 l0' For this casethe resultmust be nornralized because leadingzero is unnecessary. the S0 we mustshift thedecirnal one placeto theright to give0.9-550 l0' : 9.-550. x Noticethat

1. Select the inner cylindrical part of the center support. Be sure to select the circle and not the hatch. 2. Move the pointing device over the sidebars of the Quick Properties panel so that it expands. (Make sure the Quick Properties button on the status bar is toggled on.)

30 70 15 30 12 30 3 15 0.5 2

is based on the cd_io_errs_min metric, which is the metric that captures the rate of I/O errors on a cell disk per minute.

Let s take an example Suppose that we work out the discriminant d for a quadratic, and we find that d = 16 Then the positive square root of d is equal to j 4, and the negative square root of d is equal to j 4 Here s another example Let s revisit the quadratic stated in Practice Exercise 10 at the end of Chap 22: 4x 2 + x + 3 = 0 In this case, a = 4, b = 1, and c = 3, so d = b 2 4ac = 12 4 4 3 = 1 48 = 47 The positive square root of 47 is j (471/2), and the negative square root is j (471/2) Now let s look at the general case If d < 0, then |d | > 0.

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