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irakobra [83]
3 years ago
10

Consider three branch prediction schemes: predict not taken, predict taken, and dynamic prediction. Assume that they all have ze

ro penalties when they predict correctly and two cycles when they are wrong. Assume that the average predict accuracy of the dynamic predictor is 90%. Which predictor is the best choice for the following branches? [1.5 pts]a. A branch that is taken with 5% frequencyb. A branch that is taken with 70% frequencyc. A branch that is taken with 95% frequency
Engineering
1 answer:
SOVA2 [1]3 years ago
3 0

Answer:

Check the explanation

Explanation:

1. When a branch is taken with 5% frequency, the branch prediction scheme is "predict taken" because the process has to begin fetching and then execute at the target address.

2. When a branch is taken with 95% frequency, the branch prediction scheme is "predict not taken" because the branch outcome is definitely known.

3. When a branch is taken with 70% frequency, the prediction scheme is "dynamic prediction" the branch prediction can dynamically change during the program execution.

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prohojiy [21]

The number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

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It should be noted that when a battery is supplying power, then the positive terminal is the cathode while the negative terminal is the anode.

In conclusion, the number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

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5 0
2 years ago
A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False
tiny-mole [99]

The answer is : True

4 0
3 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
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Bas_tet [7]

Answer:

sorry i dont understand the answer

Explanation:

but i think its a xd jk psml lol

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How will you maintain the orderliness of your storage area 2pts?​
fiasKO [112]

Explanation:

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