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Elena L [17]
2 years ago
9

The frequency table shows the scores from rolling a dice 10 times.

Mathematics
2 answers:
Travka [436]2 years ago
8 0

Answer:

3.9

Step-by-step explanation:

Let me know if I'm wrong, if I am I'm sorry

Karolina [17]2 years ago
7 0
You’re gonna want to start this problem by going to your Calculator and hitting stat and then edit. Once you do that you’re going to want to type in the first column, which is one, two, three etc. Once you’re done with that column, you’re gonna want to hit your right arrow and hit the second column digits into L2. After this, you’re going to go to your stat button switch over to Calc, hit 1 – var stats, and the x with a - above it is the mean.
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F(c)= 9/5c+ 32 I just don't understand how to find the inverse of the function
Viefleur [7K]
F = 9/5 C + 32
- 32 -32

subtracting - 32 from both sides

F - 32 = 9/5c
5/9 F -32 = 5/9(9/5)C

Invert 9/5 to 5/9 multiply to both sides

5/9F - 32 = C

The answer is 5/9F-32 = C
5 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
Monique ran in a 5 kilometer race how many meters did Mo'Nique run
Blizzard [7]
5000 meter because there is 1000 meters in 1 kilometer
3 0
3 years ago
Read 2 more answers
If Sarah bought 3 pounds of apples for $6.75 and Carlos got 2 pounds of apples for $4.52,who paid less per pound and got the bet
lesantik [10]

Answer:

Sarah

Step-by-step explanation:

6.75/3 is the total for one pound . This = 2,25 while calros' 4.52/2 = 2.26. sarah paid one cent less, getting a better deal than carlos

4 0
2 years ago
The perimeter of a regular 4-sided polygon (square) is 48 units and the length of each side is x +1 units. What is the value of
iren [92.7K]

Answer:

x=11

Step-by-step explanation:

48-4=44

44/4=11

x=11

11+1=12

12•4= 48

6 0
2 years ago
Read 2 more answers
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