The question is incomplete. Here is the complete question.
A differential equation is given along with the field of problem area which it arises. Classify it as an ordinary differential equation (ODE) or a partial different equation (PDE), give the order, and indicate the independent and dependent variables. If the equation is an ordinary differential equation, indicate whether the equation is linear or nonlinear.
(aerodynamics, stress analysis)
Answer and Step-by-step explanation: The differential equation described above is anOrdinaryDifferentialEquation, because it has a definite set of variables: x and y.
It is of SecondOrder, since the highest derivative is of order 2:
The differential equation is written as derivative of a function y in terms of x, which means: IndependentVariable is X and DependentVariable is Y.
As it is an ODE, the equation is Nonlinear, because y'' or is multiplied by a variable.
This question pertains to classifying and understanding differential equations in mathematics, specifically identifying whether it is an ordinary differential equation, determining its order, and identifying independent and dependent variables.
In mathematics, a differential equation is a mathematical equation that relates a function with its derivatives. An Ordinary Differential Equation (ODE) is a differential equation containing one or more functions of one independent variable and the derivatives of those functions. The term order of a differential equation is defined as the highest power of the derivative in the equation.
As an example, if we have an equation like dy/dx = x*y, it would be an ODE since it involves only one independent variable, 'x', and the dependent variable is 'y'. The order is one since the highest order derivative is dy/dx, and the equation is nonlinear because it does not meet the criteria for a linear ODE, which stipulates that the dependent variable and its derivatives are to the first power and are not multiplied together.
#SPJ3
Answer:
It is not an empty set
Step-by-step explanation:
Obtuse angles are angles greater than 90 and less than 180.
There are quadrilaterals having 3 obtuse angles and they are possible.
If we imagine 3 obtuse angles of 91 degrees (obtuse angle), the 4th angle will be
360-91-91-91
=> 87 degrees
So, This quadrilateral can be constructed!
And also with 92, 93, 94 and so on!
So, Set A is not an empty set!
Answer:
It is not an empty set
Step-by-step explanation:
A quadrilateral with 3 obtuse angles is possible.
A obtuse angle has a measure of more than 90 degrees and less than 180 degrees.
Let’s say three angles are measuring 91 degrees in a quadrilateral.
91 + 91 + 91 + x = 360
x = 87
The measure of the fourth angle is 87 degrees which is less than 360 degrees and is a positive integer, so it is possible.
Answer:
Step-by-step explanation:
This is a test of a single population mean since we are dealing with mean.
From the information given,
Null hypothesis is expressed as
H0:μ=1.5
The alternative hypothesis is expressed as
Ha:μ>1.5
This is a right tailed test
The decision rule is to reject the null hypothesis if the significance level is greater than the p value and accept the null hypothesis if the significance level is less than the p value.
p value = 0.052
Significance level, α = 0.05
Since α = 0.05 < p = 0.052, the true statement would be
At the α=0.05 significance level, you have proven that H0 is true. B.
Answer:
There is a 69.15% probability that it weighs more than 0.8535 g.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that
The weights of a certain brand of candies are normally distributed with a mean weight of 0.8547 g, so .
We have a sample of 463 candies, so we have to find the standard deviation of this sample to use in the place of in the Z score formula. We can do this by the following formula:
Find the probability that it weighs more than 0.8535
This is 1 subtracted by the pvalue of Z when
So
has a pvalue of 0.3085.
This means that there is a 1-0.3085 = 0.6915 = 69.15% probability that it weighs more than 0.8535 g.
Can anyone actually help me work it out please
Answer:
3y=6x=18 the equation must look like this y=mx+b where m is the slope in this case it is 2 and b is the y-intercept 18 hope it helps
just divide every term by 3 to isolate y
Answer:
5.25ft
Step-by-step explanation:
i got the question right so thats it
Answer:
5 1/4
Step-by-step explanation:
5 1/4 to decimal form is 5.25