PLEASE HELP ME OUT !!!!!!!! 20 POINTSSSSSS
1 answer:
Step One
For convenience f(x) = y
y = -(1/2) sqrt(x + 3) Interchange x and y
x = -(1/2) sqrt(y + 3) Multiply both sides by -2
(-2x) = sqrt(y + 3) Square both sides
4x^2 = y + 3 Subtract 3 from both sides
4x^2 -3 = y
f-1(x) = 4x^2 - 3
Graphs
The original graph is in blue
The inverse is red.
Green Line y = x : explained below.
Comment
Inverses have a very strange property. It acts as a check.
If you perform the following operation
f-1(f(x)) you are left with x
In this case f-1[-1/2 sqrt(x + 3)]= 4(-1/2 * sqrt(x + 3))^2 - 3
f-1 (-1/2sqrt(x+3)) = 4 (1/4 (x+3) - 3
f-1(-1/2sqrt(x + 3)) = x+ 3 - 3 = x
I've included this because I think you have to put the same restriction on f-1(x) as you do on f(x) which is that for f(x) x ≥ -3
This last assertion could be incorrect, but I don't think so.
The answer is f-1(x) = 4x^2 - 3 <<<<< answer first box and
x≥-3 for the second box. <<<<<< answer for the second box.
Let me know if these are correct. I'm not sure about the restrictions.
For convenience f(x) = y
y = -(1/2) sqrt(x + 3) Interchange x and y
x = -(1/2) sqrt(y + 3) Multiply both sides by -2
(-2x) = sqrt(y + 3) Square both sides
4x^2 = y + 3 Subtract 3 from both sides
4x^2 -3 = y
f-1(x) = 4x^2 - 3
Graphs
The original graph is in blue
The inverse is red.
Green Line y = x : explained below.
Comment
Inverses have a very strange property. It acts as a check.
If you perform the following operation
f-1(f(x)) you are left with x
In this case f-1[-1/2 sqrt(x + 3)]= 4(-1/2 * sqrt(x + 3))^2 - 3
f-1 (-1/2sqrt(x+3)) = 4 (1/4 (x+3) - 3
f-1(-1/2sqrt(x + 3)) = x+ 3 - 3 = x
I've included this because I think you have to put the same restriction on f-1(x) as you do on f(x) which is that for f(x) x ≥ -3
This last assertion could be incorrect, but I don't think so.
The answer is f-1(x) = 4x^2 - 3 <<<<< answer first box and
x≥-3 for the second box. <<<<<< answer for the second box.
Let me know if these are correct. I'm not sure about the restrictions.
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Using the normal distribution, it is found that:
a) 0.8599 = 85.99% probability that x is more than 60.
b) 0.1788 = 17.88% probability that x is less than 110.
c) 0.6811 = 68.11% probability that x is between 60 and 110.
d) 0.0643 = 6.43% probability that x is greater than 125.
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- The mean is of 87, thus
.
- The standard deviation is of 25, thus
.
Item a:
This probability is <u>1 subtracted by the p-value of Z when X = 60</u>, thus:
has a p-value of 0.1401.
1 - 0.1401 = 0.8599
0.8599 = 85.99% probability that x is more than 60.
Item b:
This probability is the <u>p-value of Z when X = 110</u>, thus:
has a p-value of 0.8212.
1 - 0.8212 = 0.1788.
0.1788 = 17.88% probability that x is less than 110.
Item c:
This probability is the <u>p-value of Z when X = 110 subtracted by the p-value of Z when X = 60</u>.
From the previous two items, 0.8212 - 0.1401 = 0.6811.
0.6811 = 68.11% probability that x is between 60 and 110.
Item d:
This probability is <u>1 subtracted by the p-value of Z when X = 125</u>, thus:
has a p-value of 0.9357.
1 - 0.9357 = 0.0643.
0.0643 = 6.43% probability that x is greater than 125.
A similar problem is given at brainly.com/question/24863330