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Angelina_Jolie [31]
3 years ago
7

What is the standard deviation of scores for the tenth

Mathematics
2 answers:
Arlecino [84]3 years ago
5 0

Answer:

C

Step-by-step explanation:

GOT IT RIGHT ON TEST!!!!

ANTONII [103]3 years ago
3 0

Answer:354

Step-by-step explanation:

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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. xy = 4 (a) Find dy/d
kompoz [17]

Answer:

a) -0.9375

b) 1.25

Step-by-step explanation:

We are given the following in the question:

f(x,y) = 4

where x and y are both differentiable functions of t.

a)  x = 8 and dx/dt = 15

\\xy = 4\\\\y = \dfrac{4}{x}\\\\y = \dfrac{4}{8}=\dfrac{1}{2}\\\\\dfrac{d(f(x,y))}{dt} = 0\\\\x\dfrac{dy}{dt} + y\dfrac{dx}{dt} = 0\\\\8\dfrac{dy}{dt}  + \dfrac{1}{2}(15) = 0 \\\\\dfrac{dy}{dt} = \dfrac{1}{8}\times \dfrac{-15}{2}\\\\\dfrac{dy}{dt} = -\dfrac{15}{16}\\\\\dfrac{dy}{dt}=-0.9375

b) x = 1 and dy/dt = –5

xy = 4\\\\y = \dfrac{4}{x}\\\\ y= 4\\\\\dfrac{d(f(x,y))}{dt} = 0\\\\x\dfrac{dy}{dt} + y\dfrac{dx}{dt} = 0\\\\(1)(-5)  + 4\dfrac{dx}{dt} = 0 \\\\\dfrac{dx}{dt} = \dfrac{5}{4}\\\\\dfrac{dx}{dt} = 1.25

3 0
4 years ago
How to find slope and y intercept
Sergio [31]
Slope is rise over run
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What am i doing wrong here or am i missing something??​
Alja [10]
The answer is 7 instead of 5
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3 years ago
What's is the answer to this question: 7-6?
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Let z1 = a1 + b1i, z2 = a2 + b2i, and z3 = a3 + b3i. Prove the folowing using algebra or by showing with vectors.
Svetlanka [38]

Answer:

a)z1 +z2 =z2 + z1 ...proved.

b) z1 + ( z2+ z3 )=(z1+z2)+z3 ... proved.

Step-by-step explanation:

It is given that there are three vectors z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3. Now, we have to prove (a) z1 + z2 = z2 + z1 and (b) z1 + (z2 +z3) = (z1 + z2) + z3.

(a) z1 + z2 = (a1 +ib1) + (a2+ ib2) = (a1 +a2) + i(b1 +b2) {Adding the real and imaginary parts separately}

Again, z2 + z1 =(a2 +ib2) + (a1 +ib1) = (a2 +a1) + i(b2 +b1) {Adding the real and imaginary parts separately}

Hence, z1 +z2 =z2 + z1 {Since, (a1 +a2) = (a2 +a1) and (b1 +b2) = (b2 +b1)}

(b) z1 + ( z2+ z3 ) = [a1 + ib1] + [(a2 + a3 ) + i(b2 + b3 )] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Again, (z1+z2)+z3 = [(a1+a2) +i(b1+b2)]+[a3+ib3] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Hence, z1 + ( z2+ z3 )=(z1+z2)+z3 proved.

5 0
3 years ago
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