Answer:
And as we can see on the plot we have the distribution left skewed so then this distribution is not normal because we don't have a bell shaped histofram and is not symmetric.
Explanation:
Assuming the following dataset
Temperature (F) Frequency
50-54 2
55-59 0
60-64 4
65-69 12
70-74 7
75-79 5
80-84 1
For this case we can construct the histogram with the following R code.
f<-c(2,0,4,12,7,5,1)
> barplot(f)
And the result is on the figure attached.
And as we can see on the plot we have the distribution left skewed so then this distribution is not normal because we don't have a bell shaped histofram and is not symmetric.
To develop this problem we will apply the concept related to heat transfer defined as the product between the transfer coefficient and the temperature difference between two spaces, that is,
Here,
h = Heat transfer coefficient
= Temperature at each point
- Interior air and inner wall:
- Outer wall to exterior air:
- Inner wall to interior air:
We can see that the magnitude of the heat fluxes in the three states are the same (The negative sign only indicates the change of direction) so the wall is in steady state conditions
Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
BooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooB
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