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Amanda [17]
3 years ago
10

Please help asap i’m stupid

Mathematics
1 answer:
nevsk [136]3 years ago
3 0

Answer:

74

Step-by-step explanation:

Say that arc JL going through M is arc E and JL going the other way is arc D

For the angle formed by two tangents, K=(1/2)(E-D)

64=E-D

Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148

Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212

E is then 212

64=212-D

212-64=D=148

Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74

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Please help me answer question B!
skelet666 [1.2K]

Answer:

  • The program counter holds the memory address of the next instruction to be fetched from memory
  • The memory address register holds the address of memory from which data or instructions are to be fetched
  • The memory data register holds a copy of the memory contents transferred to or from the memory at the address in the memory address register
  • The accumulator holds the result of any logic or arithmetic operation

Step-by-step explanation:

The specific contents of any of these registers at any point in time <em>depends on the architecture of the computer</em>. If we make the assumption that the only interface registers connected to memory are the memory address register (MAR) and the memory data register (MDR), then <em>all memory transfers of any kind</em> will use both of these registers.

For execution of the instructions at addresses 01 through 03, the sequence of operations may go like this.

1. (Somehow) The program counter (PC) is set to 01.

2. The contents of the PC are copied to the MAR.

3. A Memory Read operation is performed, and the contents of memory at address 01 are copied to the MDR. (Contents are the LDA #11 instruction.)

4. The MDR contents are decoded (possibly after being transferred to an instruction register), and the value 11 is placed in the Accumulator.

5. The PC is incremented to 02.

6. The contents of the PC are copied to the MAR.

7. A Memory Read operation is performed, and the contents of memory at address 02 are copied to the MDR. (Contents are the SUB 05 instruction.)

8. The MDR contents are decoded and the value 05 is placed in the MAR.

9. A Memory Read operation is performed and the contents of memory at address 05 are copied to the MDR. (Contents are the value 3.)

10. The Accumulator contents are replaced by the difference of the previous contents (11) and the value in the MDR (3). The accumulator now holds the value 11 -3 = 8.

11. The PC is incremented to 03.

12. The contents of the PC are copied to the MAR.

13. A Memory Read operation is performed, and the contents of memory at address 03 are copied to the MDR. (Contents are the STO 06 instruction.)

14. The MDR contents are decoded and the value 06 is placed in the MAR.

15. The Accumulator value is placed in the MDR, and a Memory Write operation is performed. Memory address 06 now holds the value 8.

16. The PC is incremented to 04.

17. Instruction fetch and decoding continues. This program will go "off into the weeds", since there is no Halt instruction. Results are unpredictable.

_____

Note that decoding an instruction may result in several different data transfers and/or memory and/or arithmetic operations. All of this is usually completed before the next instruction is fetched.

In modern computers, memory contents may be fetched on the speculation that they will be used. Adjustments need to be made if the program makes a jump or if executing an instruction alters the data that was prefetched.

4 0
3 years ago
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig
Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let \mu denotes the average hours of sleep per night.

As per given , we have

H_0:\mu=7\\H_a:\mu

, since H_a is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : \overline{x}=7.24

Sample  standard deviation : s= 1.93

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

i.e.  t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58

For significance level \alpha=0.05 and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= t_{\alpha, df}=t_{0.05,21}=- 1.7207

Decision : Since the test statistic value (0.58) > critical value  1.7207, it means we are failed to reject the null hypothesis .

[Note : When |t_{cal}|>|t_{cri}|, then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

6 0
3 years ago
Five gerbils cost p dollars. How many dollars will it cost to buy 1 gerbil? How many dollars will it cost to buy g gerbils? How
Fudgin [204]
5 gerbils = p dollars
1 gerbil = p/5 dollars
g gerbils = pg / 5 dollars

5/p gerbils = 1 dollar
5d/p gerbils = d dollars
4 0
3 years ago
what TRANSFORMATIONS take place to the parent graph f(x)= log(x) to achieve the graph g(x)=log(-2x-4)+5?
Leona [35]

The -4 will translate the graph of f(x) 4 units to the right.
Then the -2 before the x will stretch it vertically with factor 2, then reflect it in the y -axis.
Finally the + 5 will translate the graph 5 units vertically upwards
4 0
3 years ago
The human eye is approximately spherical in shape with a radius of about 12 mm
Tom [10]

Answer:

2.3 cm if that's an option. Not sure tho

Step-by-step explanation:

Pls mark as brainliest

5 0
3 years ago
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