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hichkok12 [17]
3 years ago
15

Calculate 11010-1101 in base 2​

Mathematics
1 answer:
stiv31 [10]3 years ago
6 0

Answer:

1101 base 2

Step-by-step explanation:

you first take the larger number up the the lower number down

  • 11010 -1101
  • when dividing if it's zero you borrow one from the number on the left hand side
  • since it's in base 2 the number you borrow would be 2 then u subtract
  • I hoped it helped u
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Using the numbers -4, 10, 8, 2, -3, -5, write down two expressions that equal 6. You can use any relevant operation. You have to
Valentin [98]

Answer:

\sf \Bigg[ \frac{ ( - 4 \times ( - 3)) }{2} \Bigg] \: and  \: \Bigg[ \frac{ - ( - 5 \times 8)}{10} + 2 \Bigg]

Step-by-step explanation:

Given:

The \: numbers \rightarrow -4, 10, 8, 2, -3, -5

To find:

Two expressions that equal 6 using the given numbers

Solution:

Expression first,

Using numbers -4, 2, -3,

aligning the above numbers as,

\frac{ ( - 4 \times ( - 3))  }{2}

will out put 6.

<em>Verification,</em>

\frac{ ( - 4 \times ( - 3))  }{2} =  \frac{12}{2}   = \cancel\frac{12}{2} = 6

Expression second,

Using numbers 10,8,2,-5

aligning the above numbers as,

\frac{ - ( - 5 \times 8)}{10} + 2

will result 6.

<em>Verification</em>

\frac{ - ( - 5 \times 8)}{10} + 2 =  \frac{ 40}{10}  + 2   \\ \\ \frac{ - ( - 5 \times 8)}{10} + 2= 4 + 2 \\  \\\frac{ - ( - 5 \times 8)}{10} + 2  = 6

<em><u>Thanks for joining brainly community!</u></em>

5 0
2 years ago
A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi
Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

<em>Let X = number of names recalled</em>

SO, X ~ N(\mu = 19.41,\sigma^{2} = 3.17^{2})

The z-score probability distribution is given by ;

                  Z = \frac{X-\mu}{\sigma} } } ~ N(0,1)

where, \mu = mean number of names recalled = 19.41

            \sigma = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( \frac{X-\mu}{{\sigma} } } > \frac{15-19.41}{3.17}  } ) = P(Z > -1.39)

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<em>Therefore, proportion of the students recalled more than 15 names is </em><em>91.77%.</em>

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aleksandrvk [35]

Answer:

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4000 x 6.5%=260

260 x 9=2340

2340+4000=6340

8 0
3 years ago
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