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Vesna [10]
3 years ago
15

8 less than 2 times a number is greater than the sum of the same number and 9

Mathematics
1 answer:
lyudmila [28]3 years ago
4 0

step-by-step explanation:

every time we see "a number" or "the same number," we can add a variable (we'll use x here).  rewriting the problem, we get: 8 less than 2 times x is gr8er than the sum of  x and 9.  

next, we can work backwards to write an inequality.  we know from the word problem above that the number on the left side of the equation is 8 less than 2x, so we can rewrite that part like this: 2x-8.

for the right side of the equation, all we have to do is simplify "the sum of x and 9" to x + 9.  so now the inequality should look like this: 2x - 8 > x + 9.  

now we'll solve the inequality just like an equation: subtract x from both sides of the inequality, so now we have: x - 8 > 9.

then, we can add 8 to both sides to get the x by itself, and we get the answer: x > 15

hope this helps! <em>:)</em>

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How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
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Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
4_{10}=100_2\iff\log_24=2

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
8 0
3 years ago
An online data base charges $25 an hour during the day and $12.50 an hour at night. If a research company paid $250 for 12 hours
Firdavs [7]

Answer

Find out the  which system of equations could be used to determine the number of hours charged at the day rate (d) and at the night rate (n) .

To proof

As given in the question

Let us assume that the number of hours charged at the day rate = d

Let us assume that the number of hours charged at the night rate = n

An online data base charges $25 an hour during the day and $12.50 an hour at night. If a research company paid $250 for 12 hours of use.

Total working hour in a company including day and night be  12 hours .

than the equation become in the form

d + n = 12

also

An online data base charges $25 an hour during the day

$12.50 an hour at night.

If a research company paid $250 for 12 hours of use.

than the equation becomes

25d + 12.50n = 250

than the two equation are

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12.5 n = 50

n = \frac{50}{12.5}

n = 4hours

put this in the d + n = 12

d + 4 =12

d = 12-4

d = 8 hours

Therefore the number of hours charged at the day rate  is 8 hours  and at the night rate is  4hours .

Hence proved



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