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Shkiper50 [21]
3 years ago
5

Write is the base 10 number representation in 32-bit single-precision binary format (single) by 10111110110110000000000000000000

?
Mathematics
1 answer:
Bess [88]3 years ago
7 0

Answer:

(10111110110110000000000000000000)_2=(3201826816)_{10}    

Step-by-step explanation:

To find : Write is the base 10 number representation in 32-bit single-precision binary format (single) by 10111110110110000000000000000000 ?

Solution :

Converting binary format into base 10 is multiplying the digit by 2 to the place value value of digit.

Binary number is 10111110110110000000000000000000.

Converting into base 10,

=1\times 2^{31}+0\times 2^{30}+1\times 2^{29}+1\times 2^{28}+1\times 2^{27}+1\times 2^{26}+1\times 2^{25}+0\times 2^{24}+1\times 2^{23}+1\times 2^{22}+0\times 2^{21}+1\times 2^{20}+1\times 2^{19}+0\times 2^{18}+0\times 2^{17}+0\times 2^{16}+0\times 2^{15}+0\times 2^{14}+0\times 2^{13}+0\times 2^{12}+0\times 2^{11}+0\times 2^{10}+0\times 2^{9}+0\times 2^{8}+0\times 2^{7}+0\times 2^{6}+0\times 2^{5}+0\times 2^{4}+0\times 2^{3}+0\times 2^{2}+0\times 2^{1}+0\times 2^{0}

=2147483648+0+536870912+268435456+134217728+67108864+33554432+0+8388608+4194304+0+1048576+524288+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0

=3201826816

Therefore, (10111110110110000000000000000000)_2=(3201826816)_{10}

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Hans corrio 3 millas en la pista hizo un descanso y luego corio 4/5 de milla mas. Escribe el numero de millas que corrio hans co
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Answer:

19/5

Step-by-step explanation:

Aquí, queremos escribir el número total de millas que Hans corrió en forma de una fracción impropia.

Para hacer esto, agregamos todas las millas recorridas.

Matemáticamente, eso sería 3 + 4/5 = 3 4/5 Esta es la forma de fracción mixta.

La fracción impropia es así (5 * 3) + 4/5 = 19/5

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Step-by-step explanation:

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So far, Ricardo has scores of 13, 17, 19 and 21 points for the first four rounds of a dice game. What does he need the total sco
posledela

Answer:

40 points

Step-by-step explanation:

The score for the first four rounds of the game is 13, 17, 19, and 21 points.

Let S_5 and S_6 be the scores of the fifth and sixth games respectively in order to achieve an average score of 20 points per round for all six rounds.

\text{Average score} = \frac{\text{Sum of the scores of all the rounds}}{\text{Total number of the rounds}}

\Rightarrow 20=\frac{13+17+19+21+S_5+S_6}{6}

\Rightarrow 20\times6=80+S_5+S_6

\Rightarrow S_5+S_6=120-80=40.

Hence, the combined score of the fifth and sixth rounds are 40 points.

3 0
4 years ago
Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.
zysi [14]

Answer:

a) 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

b) 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d)   ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

Step-by-step explanation:

Part a

For this case we have the following data given

\bar X = 78.25 represent the sample mean for the customer order totals

\sigma =22.50 represent the population deviation

n= 40 represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be \alpha=0.1 and \alpha/2 = 0.05 and the critical value from the normal standard distirbution would be given by:

z_{\alpha/2}=1.64

And the confidence interval is given by:

\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

And replacing we got:

78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

 ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

3 0
3 years ago
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