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Crazy boy [7]
2 years ago
6

Which equation is the inverse of y = 100 – x2?

Mathematics
2 answers:
yulyashka [42]2 years ago
7 0

Answer:

The function f be described by

The graph of this function, is the graph of the parent function , which is the parabola with vertex at (0, 0) which opes upwards:

i) reflected with respect to the y-axis, because of the minus

ii) shifted 100 units up

check the picture attached.

A function has an inverse only if it is decreasing or increasing.

So we must divide the following cases for which the inverses exist:

i)  with domain (-infinity, 0]

ii)  with domain [0, infinity)

To find the formula for the inverse function g of f, we use the property:

f(g(x))=x

thus,

so each of - and + cases, are the inverses of

i)  with domain (-infinity, 0]

ii)  with domain [0, infinity)

At this point, recall that the domain of a function, is the range of it's inverse function,

and the range of the function, is the domain of the inverse function.

thus,

the range of , which is [0,infinity) is the domain of

i)  with domain [0,infinity)

So our answer is:

inverse of i)  with domain (-infinity, 0] is

and the inverse of ii)  with domain [0, infinity) is

Step-by-step explanation:

aleksandr82 [10.1K]2 years ago
4 0
The function f be described by y=100- x^{2}

The graph of this function, is the graph of the parent function x^{2}, which is the parabola with vertex at (0, 0) which opes upwards:

i) reflected with respect to the y-axis, because of the minus

ii) shifted 100 units up

check the picture attached. 


A function has an inverse only if it is decreasing or increasing.


So we must divide the following cases for which the inverses exist:

i) y=100- x^{2} with domain (-infinity, 0]

ii) y=100- x^{2} with domain [0, infinity)


To find the formula for the inverse function g of f, we use the property:

f(g(x))=x

thus, 

f(g(x))=x\\\\ 100-[g(x)]^2=x\\\\g^2(x)=100-x\\\\g(x)= \mp\sqrt{100-x}



so each of - and + cases, are the inverses of 


i) y=100- x^{2} with domain (-infinity, 0]

ii) y=100- x^{2} with domain [0, infinity)


At this point, recall that the domain of a function, is the range of it's inverse function, 

and the range of the function, is the domain of the inverse function.

thus, 

the range of \sqrt{100-x}, which is [0,infinity) is the domain of 

i) y=100- x^{2} with domain [0,infinity)



So our answer is:

inverse of i) y=100- x^{2} with domain (-infinity, 0] is -\sqrt{100-x}


and the inverse of ii) y=100- x^{2} with domain [0, infinity) is 

\sqrt{100-x}

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If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen
Lunna [17]
Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0
2 years ago
Which statement is true about the end behavior of the graphed function?
Tems11 [23]

Answer:

(SEE EXPLANATION FIRST)

Answer D is TRUE because on the graph if you were to keep going left even after the picture of the graph ends, the graph will still keep going up infinite

Step-by-step explanation:

If this is the graph you can read the answer

6 0
2 years ago
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Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
Firlakuza [10]

Answer:

the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

n/4 = E(X1²) - (n/2)²

E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
1 year ago
In a study in Scotland (as reported by Devlin 2009), researchers left a total of 320 wallets around Edinburgh, as though the wal
Ann [662]

Answer:

a) The observed proportion of wallets that were returned

  p = 0.45625

b) <em> 95% of confidence intervals for Population proportion</em>

<em>  0.40168  , 0.51082)</em>

<em>c) The lower bound of the 95% confidence interval = 0.40168</em>

<em>d) The upper bound of the 95% confidence interval = 0.51082</em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

a)

Given data the  researchers left a total of 320 wallets around Edinburgh, as though the wallets were lost. Each contained contact information including an address. Of the wallets, 146 were returned by the people who found them

Given sample size 'n' = 320

  Given data          'x ' = 146

<em>Sample proportion </em>

              p = \frac{x}{n}

             p = \frac{x}{n} = \frac{146}{320} = 0.45625

<u><em>Step(ii)</em></u>:-

b)<em> </em><u><em>95% of confidence intervals for Population proportion</em></u>

Level of significance = 95% or 0.05%

Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96

<em>95% of confidence intervals for Population proportion are determined by</em>

<em></em>(p - Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} } , p + Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} })<em></em>

<em></em>(0.45625 - 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} } , 0.45625 + 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} })<em></em>

<em>(0.45625 - 0.05457 , 0.45625 + 0.05457)</em>

<em>(   0.40168  , 0.51082) </em>

<em>c) The lower bound of the 95% confidence interval = 0.40168</em>

<em>d) The upper bound of the 95% confidence interval = 0.51082</em>

7 0
2 years ago
A customer/member owes $11.02 and pays $100.00 in cash. Change due is $88.98. True or False?
Ugo [173]

Answer:

  true

Step-by-step explanation:

$100 -11.02 = $88.98 . . . . true

3 0
2 years ago
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