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Tcecarenko [31]
3 years ago
15

Suppose that a teacher plans to give four students a quiz. The minimum possible score on the quiz is 0, and the maximum possible

score is 10.
What is the set of four student scores that would make the standard deviation as large as it could possibly be? Use your calculator to find this largest possible standard deviation.
Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

5 is the largest it could possibly be

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Anyone have any idea what the pattern may be? Need help ASAP!
Anna007 [38]

Answer:

21, 34, 55

Step-by-step explanation:

13+8=21

21+13=34

34+21=55

4 0
3 years ago
Write an equivalent form of the exponential<br> expression:<br> (x4)3
ELEN [110]

Answer:

x¹²

Step-by-step explanation:

You need to indicate which terms are exponents. For example, (x⁴)³ can be written as (x^4)^3.

(x⁴)³ = x⁴*³ = x¹²

5 0
2 years ago
it takes 18 3/8 inches of wood to make a frame for a small picture if Ms Jones has 108 inches of wood how many frames can she ma
scZoUnD [109]
It would make 5 complete frames, and about 8/10 of another frame.
108 / 18.375 = 5.8&*)&^&%*$*
I used a calculator.
You could keep adding 18 and 3/8 over and over again until you get right under 108 if you're good at math like me, but most people would have to add 18 and 3/8 separately.

Thanks, and if you liked my answer, please consider giving me the brainliest answer, because I need 5 of them to get to the next level.
6 0
3 years ago
Read 2 more answers
The area of the trapezoid is 70 square inches. What is the length of the larger base?
GalinKa [24]

Answer:

20 i think srry if wrong

Step-by-step explanation:

5 0
3 years ago
Express p in terms of q if
Irina-Kira [14]

p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)

q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}

8 0
2 years ago
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