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Jlenok [28]
3 years ago
11

The area of a square is 1,000 ft2. About how long is one side of the square?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
7 0
500 feet. Hope this helped
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PLEASE HELP AS SOON AS POSSIBLE
frutty [35]

Answer:

FALSE

TRUE

TRUE

Step-by-step explanation:

For median,

Arrange your numbers in numerical order.

Count how many numbers you have.

If you have an odd number, divide by 2 and round up to get the position of the median number.

If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.

Source: https://www.verywellmind.com/how-to-identify-and-calculate-the-mean-median-or-mode-2795785

So for April median: 2.5, 3, 3.5, 3.5 (since we have 4 numbers we divide by 2 and we count by 2 places from left. Since we have even number data, we average 3 with the next higher position to get the median so (3+3.5)/2 = 3.25

For May apply same concept and you get median to be 2.25.

Median difference is 3.25-2.25 = 1

Therefore, statement 1 is false and statement 2 is true.

For average in May, add all numbers and divide by the number of data points so May= (2.5+3+3.5+3.5)/ 4 = 3.13

Apply same concept you for April and you get 2.38 as mean.

Mean difference is 3.13-2.38 = 0.75

Therefore, statement 3 is correct (true)

4 0
3 years ago
Lolz please help me I would gladly appreciate it
Tanzania [10]

Pentagon has sum of 540°

6 0
3 years ago
Read 2 more answers
What is 6 1/2 + 10? please tell me and explain for me to understand
ipn [44]

Answer:

33/2

Step-by-step explanation:

6 1/2 ---> 13/2

10 ---> 20/2

13/2 + 20/2 =

33/2

5 0
3 years ago
Simplify each expression. -4x + 1 + 6 - (-9x)
IceJOKER [234]

Step-by-step explanation:

-4x+1+6-(-9x)

-4x+7-(-9x)

5x+7

7 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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