Answer:
a. E(x) = 3.730
b. c = 3.8475
c. 0.4308
Step-by-step explanation:
a.
Given
0 x < 3
F(x) = (x-3)/1.13, 3 < x < 4.13
1 x > 4.13
Calculating E(x)
First, we'll calculate the pdf, f(x).
f(x) is the derivative of F(x)
So, if F(x) = (x-3)/1.13
f(x) = F'(x) = 1/1.13, 3 < x < 4.13
E(x) is the integral of xf(x)
xf(x) = x * 1/1.3 = x/1.3
Integrating x/1.3
E(x) = x²/(2*1.13)
E(x) = x²/2.26 , 3 < x < 4.13
E(x) = (4.13²-3²)/2.16
E(x) = 3.730046296296296
E(x) = 3.730 (approximated)
b.
What is the value c such that P(X < c) = 0.75
First, we'll solve F(c)
F(c) = P(x<c)
F(c) = (c-3)/1.13= 0.75
c - 3 = 1.13 * 0.75
c - 3 = 0.8475
c = 3 + 0.8475
c = 3.8475
c.
What is the probability that X falls within 0.28 minutes of its mean?
Here we'll solve for
P(3.73 - 0.28 < X < 3.73 + 0.28)
= F(3.73 + 0.28) - F(3.73 + 0.28)
= 2*0.28/1.3 = 0.430769
= 0.4308 -- Approximated
Answer:
Step-by-step explanation:
Given
Ferret:
Labrador Retriever
Required
Determine the slope
Represent the given as x or y
Taking the weight as explanatory variable;
We have:
x = 2.1; y = 3.4
x = 7.5; y = 70
Slope is calculated as:
<em>Hence, the slope is 12.33</em>
If I'm understanding your question, the drawing is 10cm and for every 10cm, the drawing is etched 1mm. This means that no matter how big the drawing gets, the actual object will always be bigger since it's 10cm larger always than the drawing. Hope this helped!
Answer: 301mm^3
Step-by-step explanation:
V=πr^2h/3
volume= 3.14 * 6^2 * 8/3
301mm^3
Just go to were the line intercepts the y axis so -2 and rise over run so 2 up and 2 over but its negative so C should be correct.
