Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s
emester I-2018? List the names of the students who have taken at least one course that Professor Collins is qualified to teach. List the names of the students who took at least one course with ""Syst"" in its name during the semester I-2018.
1 answer:
Answer:
See explanations
Explanation:
Consider the following Venn diagram to retrieve the name of course that student Altvater took during semester I-2015.
• In above Venn diagram inner Ellipse represent the subquery part, this subquery part select the Student ID from STUDENT table.
• Second sub query is used to determine the SectionNo of all student whose studentID retrived in the first subquery
• Finally the main query displays the CourseName from COURSE table.
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Answer:
1.505
Explanation:
cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
stress is force per unit area
stress=P/A
A = πd^2/4.
uncertainty of axial force P= +/-.11
s=+/-.20, strength
d=+/-.04 diameter
fail load/max allowed
minimum design=fail load/max allowed
minimum design =s/(P/A)
sA/P
A=(.96d^2)/4, so Amin=
(because the diameter at minimum is (1-0.04=0.96)
minimum design=Pmax/(sminxAmin)
1.11/(.80*.96^2)=
1.505
The total amount of daily heat transfer is 1382.38 M w.
The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.
<u>Explanation:</u>
Given data,
= 10° C
= 250 w/
k
Pipe length = 20 m
Inner diameter = 6 cm,
= 3 cm
Outer diameter = 8 cm,
= 4 cm
The thickness of insulation is 4 cm.
=
+ 4
= 4+4
= 8 cm
is the heat transfer coefficient of convection inside,
is the heat transfer coefficient of convection outside.
The heat transfer rate between ambient and steam is
watt
= watt
= watt
q = 15999.86 watt
The total amount of daily heat transfer = 15999.86 × 86400
= 1382.387904 watt
= 1382.38 M w
The total amount of daily heat transfer is 1382.38 M w.
b) The temperature on the outside surface of the gypsum plaster insulation.
q =
15999.86
- 10 = 7.96
= 17.96 ° C.
Answer:
T_{f} = 90.07998 ° C
Explanation:
This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process
Qc = -Qb
M (T₁ -
) = m
(T_{f}-T₀)
Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³
ρ = m / V
V = 1000 -100 = 900 ml
V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³
V_air = 0.100 l = 0.1 10⁻³ m³
Tea Mass
M = ρ V_te
M = 1100 0.9 10⁻³
M = 0.990 kg
Air mass
m = ρ _air V_air
m = 1.225 0.1 10⁻³
m = 0.1225 10⁻³ kg
(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀
T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})
Let's calculate
T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)
T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)
T_{f} = 98097.11 / 1089.0002
T_{f} = 90.07998 ° C
This temperature decrease is very small and cannot be measured
Answer:
Now we can decide based on the significance level . If
we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.
we see that
so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
Explanation:
For this case they conduct the following system of hypothesis for the ture mean of interest:
Null hypothesis:
Alternative hypothesis:
The statistic for this hypothesis is:
And on this case the value is given
For this case in order to take a decision based on the significance level we need to calculate the p value first.
Since we have a lower tailed test the p value would be:
Now we can decide based on the significance level . If
we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.
we see that
so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15