Answer:
The length of the field on the drawing is 55 cm.
Step-by-step explanation:
Given:
Mark made a scale drawing of a soccer field.
Using a scale of .5 cm=1 m.
The actual length of the field is 110 m.
Now, to find the length of the field on drawing.
Let the length of the field on drawing be 
As given 0.5 cm is equivalent to 1 m.
Thus, 
is equivalent to 110 m.
Now, to get the length of the field on drawing by using cross multiplication method:
<em>By cross multiplying we get:</em>
⇒ 
⇒ 
Therefore, the length of the field on the drawing is 55 cm.
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
a. 
has an average value on [5, 11] of
b. The mean value theorem guarantees the existence of 
such that 
. This happens for
Answer:
675
Step-by-step explanation:
use the legend under the map
