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nalin [4]
2 years ago
9

Write each of the following numbers as a power with an exponent to the base 2:

Mathematics
2 answers:
sergejj [24]2 years ago
5 0

Powers of 2 are ...

... (x, 2^x) ∈ {(0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32), (6, 64), (7, 128), (8, 256), (9, 512), (10, 1024)}

By the rules of exponents, ...

... 1/a^b = a^-b

Using the above table, ...

a) 1 = 2^0

b) 1/4 = 2^-2

c) 1/64 = 2^-6

d) 1/256 = 2^-8

e) 64 = 2^6

f) 1/16 = 2^-4

g) 256 = 2^8

h) 1/1024 = 2^-10

Alika [10]2 years ago
5 0

Answer:

a) 1  = 2^0

b) 1/4  = 2^-2

c) 1/64  =  2^-6

d) 1/256  =  2^-8

e) 64  = 2^6

f) 1/16  = 2^-4

g) 256  = 2^8

h) 1/1024 = 2^-10

Step-by-step explanation:

1 /a^b = a^ -b   and a^0 =1

a) 1  = 2^0

b) 1/4  = 1/2^2 = 2^-2

c) 1/64  = 1/ 2^6 = 2^-6

d) 1/256  = 1/2^8 = 2^-8

e) 64  = 2^6

f) 1/16  =1/2^4 = 2^-4

g) 256  = 2^8

h) 1/1024 =1/ 2^10  = 2^-10

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Kendle wants to play several games of laser tag. She has $35 to play g games. Each
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2 years ago
Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat
anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

 M be the event of failure due to mechanical defects, P(M) = 0.88

 LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

 IA be the event of mechanical defect due to improper assembly, P(IA/M)   =0.73

 DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

 IC be the event of electrical connects due to improper connections,

  P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

  P(PWW/F) = 0.52

(b)                                     <u> </u><u>Keyboard failures</u>

<h2>                              /               \</h2>

           <u> </u><u> Faulty electrical connects   </u>            <u>Mechanical Defects </u>          

                      P(F) = 0.12                                             P(M) = 0.88

<h2>        /            |             \                  /            \</h2>

<u><em>Defective wires</em></u>  <u><em>Improper</em></u>        <u><em>Poorly</em></u>                  <u><em>Loose Keys</em></u>      <u><em>Improper</em></u><em> </em>

P(DW/F)=0.35   <u><em>Connections</em></u>   <u><em>Welded wires</em></u>      P(LK/M)=0.27   <em> </em><u><em>Assembly</em></u>

                           P(IC/F)=0.13     P(PWW/F)=0.52                            P(IA/M)=0.73              

This is the required tree diagram.

(c) Probability that a failure is due to loose keys is given by:

  P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose  

                                               keys}

    P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

     wires is given by P(IC \bigcup PWW) ;

 P(IC \bigcup PWW) = P(IC) + P(PWW) - P(IC \bigcap PWW) { Here P(IC \bigcap PWW) = 0 }

 P(IC) = P(IC/F) * P(F)  = 0.13 * 0.12 = 0.0156

 P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC \bigcup PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

8 0
3 years ago
pam deposited $6,900 in a savings account with simple interest. Four months later she had earned $23 in interest. What was the i
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Answer:

1%

Step-by-step explanation:

We use the simple interest equation of A=P(1+rt).

A-the total amount with interest earned

P-the initial amount or principal

r-rate

t-time in years

We substitute the values P=6900, A=6923, and t=0.33 since 4 months divided by 12 months is 0.333 years. We then solve for r.

6923=6900(1+r(0.33))\\6923=6900(1+0.33r)\\\frac{6923}{6900} =\frac{6900(1+0.33r)}{6900} \\1.0033=1+0.33r\\1.0033-1=1-1+0.33r\\0.0033=0.33r

Our final step is to divide both sides by 0.33.

\frac{0.0033}{0.33} =\frac{0.33r}{0.33} \\0.01=r

This is the decimal of the rate. We convert to a percentage by multiplying by 100. 0.01(100)=1%.


5 0
2 years ago
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