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enot [183]
3 years ago
8

Determine the inverse function of f(x)=

{4x-1} " alt=" \sqrt{4x-1} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
gladu [14]3 years ago
7 0
f(x)=\sqrt{4x-1}\\
y=\sqrt{4x-1}\\
y^2=4x-1\\
4x=y^2+1\\
x=\dfrac{y^2+1}{4}\\
f^{-1}(x)=\dfrac{x^2+1}{4}\\
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Find the domain of the following expression 15 - √(x+2)
ipn [44]

Answer:

[-2, ∞]

Step-by-step explanation:

15 - √(x+2)

domain is any value as long as (x+2) is not-negative, since √ of a negative number has no Real solution.

x+2 ≥ 0 ⇒ x ≥ -2

6 0
2 years ago
I just started 8th grade and I want to know if my answer is right.
mario62 [17]
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8 0
3 years ago
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To win a bowling trophy you need a three-game total score of at least five hundred on the first two games your scores are 183 an
pickupchik [31]
We can write an equation for this problem being, 500-(183+165)=152. So the answer is that your next must have a score of at least 152 or higher in order to win a bowling trophy.
7 0
3 years ago
8. An intrusion prevention system can either wait until it has all the information it needs or can allow packets through based o
MrRa [10]

<u>Explanation:</u>

<em>Remember, </em>any network security system that constantly monitors a network infrastructure in other to detect and prevent threats to the network is called an intrusion prevention system (IPS.

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4 0
2 years ago
There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=
yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = \frac{12P4}{12^{4} }

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = \frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }

P(X=2) = 0.401

P(X=3) = \frac{4C3 * 12P2}{12^{4} }

P(x=3) = 0.025

P(X=4) = \frac{12}{12^{4} }

P(X=4) = 0.001

Mean, \mu = \sum xP(x)

\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879

Standard deviation, SD = \sqrt{\sum x^{2} P(x) - \mu^{2}}  \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2}  * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}

SD = 1.033

4 0
3 years ago
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