Answer:
The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.
Step-by-step explanation:
Let the random variable <em>X</em> denote the water depths.
As the variable water depths is continuous variable, the random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 2.00 m and <em>b</em> = 7.00 m.
The probability density function of <em>X</em> is:
Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:
![=\frac{1}{5.00}\int\limits^{5.00}_{2.25} {1} \, dx\\\\=0.20\times [x]^{5.00}_{2.25} \\\\=0.20\times (5.00-2.25)\\\\=0.55](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cint%5Climits%5E%7B5.00%7D_%7B2.25%7D%20%7B1%7D%20%5C%2C%20dx%5C%5C%5C%5C%3D0.20%5Ctimes%20%5Bx%5D%5E%7B5.00%7D_%7B2.25%7D%20%5C%5C%5C%5C%3D0.20%5Ctimes%20%285.00-2.25%29%5C%5C%5C%5C%3D0.55)
Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.
Answer:
i. LM || NO (converse alternate interior angle theorem)
ii. <1 ≅ <2 (alternate interior angle theorem)
Step-by-step explanation:
Two or more lines are said to be parallel if they do not meet when extended, even till infinity.
Alternate angles are said to be equal in measure.
Given that;
<1 ≅ < 3,
Since <2 ≅ <3 (alternate interior angle theorem)
Then,
<1 ≅ <2 (transitive property)
Also,
<1 ≅ <2 (alternate interior angle theorem)
Therefore since <1 ≅ <2, thus;
LM || NO (converse alternate interior angle theorem)
Answer:
a.Z(-2,1)
b.Z(1,1)
c.Z(-3,2)
Step-by-step explanation:
z(-2,3)
Imagine this point on a graph.
Translate it down two units :
the x stays -2, by going down the y decreases 2 so 3-2=1
Z(-2,1)
Translate Right three units : I'm assuming that we use the answer from the first translation
Z(-2,1)
The y doesn't change this time the x increases 3 since we're moving to the right.
Z(1,1)
Translate up 1 and left 4:
Z(1,1)
by moving up one we have Z(1,2) then by moving 4 to the left we get Z(-3,2)
Hope this helps :)
Answer:
A.) gf(x) = 3x^2 + 12x + 9
B.) g'(x) = 2
Step-by-step explanation:
A.) The two given functions are:
f(x) = (x + 2)^2 and g(x) = 3(x - 1)
Open the bracket of the two functions
f(x) = (x + 2)^2
f(x) = x^2 + 2x + 2x + 4
f(x) = x^2 + 4x + 4
and
g(x) = 3(x - 1)
g(x) = 3x - 3
To find gf(x), substitute f(x) for x in g(x)
gf(x) = 3( x^2 + 4x + 4 ) - 3
gf(x) = 3x^2 + 12x + 12 - 3
gf(x) = 3x^2 + 12x + 9
Where
a = 3, b = 12, c = 9
B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)
To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula
Y = 3x + 3
X = 3y + 3
Make y the subject of formula
3y = x - 3
Y = x/3 - 3/3
Y = x/3 - 1
Therefore, g'(x) = x/3 - 1
For g'(12), substitute 12 for x in g' (x)
g'(x) = 12/4 - 1
g'(x) = 3 - 1
g'(x) = 2.
