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GalinKa [24]
3 years ago
5

If you vertically compress the square root parent function f(x) = ^x, by 1/2 of a unit what is the equation of the new function

Mathematics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

The equation of the new function is g(x) = \frac{\sqrt{x}}{2}

Step-by-step explanation:

Suppose we have a function f(x).

a*f(x), a > 1, is vertically stretching f(x) a units. Otherwise, if a < 1, we are vertically compressing f(x) by a units.

f(x - a) is shifting f(x) a units to the right.

f(x + a) is shifting f(x) a units to the left

In this question:

f(x) = \sqrt{x}

Vertically compressing by 1/2:

This is the same as multiplying the function by 1/2. So

\frac{1}{2} \times \sqrt{x} = \frac{\sqrt{x}}{2}

The equation of the new function is g(x) = \frac{\sqrt{x}}{2}

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Shtirlitz [24]

Answer:

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Step-by-step explanation:

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3 years ago
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Jlenok [28]

Answer:

a) 0.9641.

b) 0.0082

c) 0.0277

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

(a) ...to the left of 1.8.

p-value of Z = 1.8, which, looking at the z-table, is of 0.9641.

(b) ...to the right of 2.4.

1 subtracted by the p-value of Z = 2.4.

Looking at the z-table, Z = 2.4 has a p-value of 0.9918.

1 - 0.9918 = 0.0082, which is the answer.

(c) ...between 1.8 and 2.4.

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6 0
3 years ago
Determine the commutators of the operators a and a+,where a = (x + ip)/2 ^1/2 and a+ = (x - ip)/ 2 ^1/2
Vsevolod [243]

Answer:

Given that:

a = (\frac{x+ip}{2})^{\frac{1}{2}} and a+= (\frac{x-ip}{2})^{\frac{1}{2}}

if a , a+ commutator, it obeys aa^+ = a^+a

First find:

aa^+ = (\frac{x+ip}{2})^{\frac{1}{2}} (\frac{x-ip}{2})^{\frac{1}{2}}

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Now;

a^+a =(\frac{x-ip}{2})^{\frac{1}{2}} (\frac{x+ip}{2})^{\frac{1}{2}} = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}

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therefore, aa^+ = a^+a which implies the operators a and a+ are commutators.    


7 0
3 years ago
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