Answer:
Amount of money Judy has in her account is
.
Amount of money Tommy has in his account is 
Step-by-step explanation:
To find : Amount of money in Judy account and amount of money in Tommy account:
Solution:
Given:
Judy has $7500 more than a number in her account.
Let the number be 'n'.
So we can say that;
Amount of money in Judy account is equal to 7500 plus number.
framing in equation form we get;
Amount of money in Judy account = 
Hence Amount of money Judy has in her account is
.
Now Given:
Tommy has 6 more than 3 times the amount of money in his bank account than Judy.
So we can say that;
Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.
framing in equation form we get;
Amount of money in Tommy account = 
Hence Amount of money Tommy has in his account is
.
Answer:

Step-by-step explanation:
Given: 76,45,64,80,92
Required: Determine the standard deviation
We start by calculating the mean

Where x-> 76,45,64,80,92 and n = 5



Subtract Mean (71.4) from each of the given data

Determine the absolute value of the above result

Square Individual Result

Calculate the mean of the above result to give the variance


Hence, Variance = 255.298
Standard Deviation is calculated by 



Answer with Step-by-step explanation:
We are given that six integers 1,2,3,4,5 and 6.
We are given that sample space
C={1,2,3,4,5,6}
Probability of each element=
We have to find that 
Total number of elements=6
={1,2,3,4}
Number of elements in
=4

Using the formula

={3,4,5,6}
Number of elements in
=4

={3,4}
Number of elements in 

{1,2,3,4,5,6}

Answer:
2
Offmind's Step-by-step explanation:
The Mode of a data set (<em>line plot in this case</em>) is the number which occurs most often. The line plot here shows the most x's on the number 2 than any other number so the mode is in fact, 2.
Bonus Fact: In another case, if there had been two numbers with the same amount of x's then those would both be considered the mode!
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-<em>Offmind</em>
4 meters per second. you just divide 28 seconds with 12 meters and that will give you 4 meters per second.Please mark as brainliest...