We are given dimensions of model and actual figure.
model: 9.5cm actual : 30.5m.
We need to find the length of actual shape for the scale factor 5cm.
Let us assume actual length be x m.
Let us set a proportion now.
<h3>5 : x = 9.5 : 30.5</h3>
Let us convert proportion into fractions.
On cross multiplying, we get
9.5x = 5× 30.5
9.5x = 152.5.
On dividing both sides by 9.5, we get
x=16.05.
<h3>Therefore, the scale factor for a model is 5cm= 16.05 m.</h3>
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:
<h2>x = 28</h2><h2>CE = 132</h2><h2 />
Step-by-step explanation:
|<------------- CE -------------------->|
C-------------------D-------------------E
2x + 10 3x - 18
2x + 10 = 3x - 18
2x - 3x = -18 - 10
-x = -28
x = 28
CE = 2x + 10 + 3x - 18
CE = 2(28) + 10 + 3(28) - 18
CE = 56 + 10 + 84 - 18
CE = 132
We know that the area is 20.
Since 20 is a small number: lets list out possible combinations of lengths and widths.
1 * 20
2 * 10
4 * 5
L = 7 + 3w
lets see which on makes sense.
L = 7 + 3w
20 = w7 + 3w^2
3w^2 + 7w -20 = 0
(3w - 10)(w - 2)
w can equal 10/3 or 2.
So the dimensions: are Width = 2 Length = 10
