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Serjik [45]
2 years ago
8

Question 12(Multiple Choice Worth 1 points)

Mathematics
1 answer:
rusak2 [61]2 years ago
7 0

Answer:

\frac1{25}

Step-by-step explanation:

Hello!

Let's use a couple of exponent rules:

  • \frac{a^x}{a^b} = a^{x - b}
  • a^{-x} = \frac{1}{a^x}

Let's utilize the first rule and simplify the fraction.

<h3>Simplify</h3>
  • \frac{5^4}{5^6}
  • 5^{4 - 6}
  • 5^{-2}

Now, we can utilize the second exponent rule, as we have a base to the power of a negative number.

<h3>Simplify</h3>
  • 5^{-2}
  • \frac1{5^2}
  • \frac1{25}

The answer is option 1, \frac{1}{25}.

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Determine if the following scale factor would be an enlargement or reduction.<br> scale factor: 2.5
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Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i
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i. The Lagrangian is

L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)

with critical points whenever

L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4

L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

L_\lambda=4x^2+y^2-9=0

  • If x=0, then L_\lambda=0\implies y=\pm3.
  • If y=0, then L_\lambda=0\implies x=\pm\dfrac32.
  • Either value of \lambda found above requires that either x=0 or y=0, so we get the same critical points as in the previous two cases.

We have f(0,-3)=9, f(0,3)=9, f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25, and f\left(\dfrac32,0\right)=\dfrac{729}4, so f has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)

with critical points whenever

L_x=2\lambda x=0\implies x=0 (because we assume \lambda\neq0)

L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda

L_\lambda=x^2+y^2+z^2-36=0

  • If x=y=0, then L_\lambda=0\implies z=\pm6.
  • If \lambda=-1, then z=-5, and with x=0 we have L_\lambda=0\implies y=\pm\sqrt{11}.

We have f(0,0,-6)=60, f(0,0,6)=-60, f(0,-\sqrt{11},-5)=61, and f(0,\sqrt{11},-5)=61. So f has a maximum value of 61 and a minimum value of -60.

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3 years ago
A researcher conducts a hypothesis test using a sample of n = 40 from an unknown population. What is the df value for the t stat
aliya0001 [1]

Answer:

Correct option: b 39.

Step-by-step explanation:

A one-sample (or one mean) t-test is applied to test whether the population-parameter is significantly different from some hypothetical value.

The degrees of freedom of the <em>t</em>-test is:

df = n - 1

It is provided that the sample size is, <em>n</em> = 40.

Compute the degrees of freedom as follows:

df = n - 1

   = 40 - 1

   = 39

Thus, the correct option is b.

3 0
3 years ago
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