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diamong [38]
3 years ago
14

Lily is at buckingham fountain (5,5) and wants to meet david, who is at millennium park (2,1) for lunch. They want to meet at th

e landmark that is closest to the midpoint of their location. At which landmark should they meet?
Mathematics
1 answer:
Nat2105 [25]3 years ago
4 0
(3.5,3.5) because then you would just find the midpoint in between the two point.
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Explain the relationship between multiplying and diving by 10, 100, and 1,000 and moving the decimal point to the right or to th
zhannawk [14.2K]
If you're multiplying by 10, you're moving the decimal point to the right, the same with any multiplication in the 10's, so 100, 1,000, etc. This is because you are making the number 10 times larger, so if you had 3.4, the 4 is in the 1/10 column so multiplying by 10 would put it in the units column, making the number 34.

On the other hand, dividing by the 10's makes the decimal point move to the left because you are making the number 10 times smaller, so if the units column is worth 1, then 10 times less than that is 1/10, which is the first decimal place.

I hope this helps! I tried to explain it well but let me know if you still don't understand or if I've confused you in any way :)
5 0
3 years ago
What is 12.81 rounded to the tens place
nekit [7.7K]

Answer: rounded to the tens is 10 the tens place is the one in 12.81 or like the one in 13.81

Step-by-step explanation:

6 0
3 years ago
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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b
jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that  R should represent the set of all possible outcomes generated from  a bit string of length 10 .

So; as each place is fitted with either 0 or 1

\mathbf{|R|= 2^{10}}

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if  a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

\mathbf{P(E) = \dfrac{|E|}{|R|}}

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

\mathbf{{|E|}=1 }   ; \mathbf{|R|= 2^{10}}

\mathbf{P(E) = \dfrac{1}{2^{10}}}

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3 years ago
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coldgirl [10]
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6 0
2 years ago
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Help please my moms going to kill me if I don’t get this right
iragen [17]

Answer:

x=-\frac{41}{6}

Step-by-step explanation:

Pay attention to the two tick marks. This indicates that those sides are congruent to each other and so will the angles on the bottom and top. Therefore, we can set up the equation 137+2(-3x+1)=180 and solve for x:

137+2(-3x+1)=180

137+(-6x+2)=180

137-6x+2=180

139-6x=180

-6x=41

x=-\frac{41}{6}

8 0
2 years ago
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