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My name is Ann [436]
3 years ago
6

Suki split five dog treats equally among her six dogs which fraction represents this division

Mathematics
1 answer:
aleksley [76]3 years ago
6 0

1/6 of a treat would be your answer

hope this helps

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What is 9 percent of 397? And rate my dog 1/100 her name is Sophie, Soph for short!
Deffense [45]

Answer:

Omg 1000/10 for soph. and 35.73

Step-by-step explanation:

7 0
3 years ago
A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who
dezoksy [38]

Answer:

The sample  size is  n =68

Step-by-step explanation:

    The population proportion is  \^ p =  0.50

     The margin of error is  E =  0.1

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>  n = [\frac{1.645 }{0.1} ]^2 * 0.50  (1 - 0.50 )

=>  n =68

     

6 0
3 years ago
How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
ad-work [718]
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
What's 0 divided by 0?
Vitek1552 [10]

Answer:

Nothing

Step-by-step explanation:

You can't divide by zero

7 0
3 years ago
Read 2 more answers
Solve n + 5(n - 1) = 7
Nina [5.8K]

Answer:

N = 2

Step-by-step explanation:

n + 5(n-1) = 7

n + 5n - 5 = 7

6n - 5 = 7

6n = 12

n = 2

HOPE THIS HELPS

PLZZ MARK BRAINLIEST!!!

7 0
3 years ago
Read 2 more answers
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