The new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.
According to the question,
Original mean is 10 and original standard deviation is 5 . In order to find to new mean and standard deviation when each score in data set is multiplied by 5 and then 7 is added.
First "change of scale" when every score in a data set is multiplied by a constant, its mean and standard deviation is multiplied by a same constant.
Mean: 10*3 = 30
Standard deviation: 5*3 = 15
Secondly "change of origin" when every score in a data set by a constant, its mean get added or subtracted by the same constant and standard deviation remains constant.
Applying change of origin in the above mean and standard deviation
Mean: 30 - 4 = 26
Standard deviation: Remains same = 15
Hence, the new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.
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Answer:
124
Step-by-step explanation:
18+36=54
i tried to do all of these and none of them add up to 54
Answer and Step-by-step explanation:
If the radius is 5, then we can plug 5 into the area of a circle equation.
A = 
A = 
A = 25
<u>The area of a circle with radius 5 is 25</u><u>.</u>
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<span>The number of x-intercepts that appear on the graph of the function
</span>f(x)=(x-6)^2(x+2)^2 is two (2): x=6 (multiplicity 2) and x=-2 (multiplicity 2)
Solution
x-intercepts:
f(x)=0→(x-6)^2 (x+2)^2 =0
Using that: If a . b =0→a=0 or b=0; with a=(x-6)^2 and b=(x+2)^2
(x-6)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x-6)^2] = sqrt(0)→x-6=0
Adding 6 both sides of the equation:
x-6+6=0+6→x=6 Multiplicity 2
(x+2)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x+2)^2] = sqrt(0)→x+2=0
Subtracting 2 both sides of the equation:
x+2-2=0-2→x=-2 Multiplicity 2
