Given z=f(x,y),x=x(u,v),y=y(u,v), with x(1,3)=2 and y(1,3)=2, calculate zu(1,3) in terms of some of the values given in the tabl
stich3 [128]
The value of zu(1,3) using the data elements represented on the table of values is q + p
<h3>How to solve the calculus expression?</h3>
The given parameters are:
z = f(x, y)
x = x(u, v)
y = y(u, v)
Where
x(1, 3) = 2 and y(1, 3) = 2
To calculate zu(1,3), we make use of:
The values x(1, 3) = 2 and y(1, 3) = 2 mean that:
(x,y) = (2,2).
So, we have:
From the table of values, we have:
So, the equation becomes
Evaluate the product
Hence, the value of zu(1,3) is q + p
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Answer:
The enrollment after 5 years is 10,724
Step-by-step explanation:
Generally, we can have the depreciation formula written as follows;
A = P(1 - r)^t
A is the number of enrollment in after a certain number of years t
P is the initial population which is 13,500
r is the rate of depreciation which is 4.5% = 4.5/100 = 0.045
t = 5 years
Substituting these values, we have it that;
A = 13,500(1-0.045)^5
A = 10,723.84
He is scared and is running away from his past
Answer:
16
Step-by-step explanation:
First, substitute the values of the variables into the expression:
3(y + 4) - (x - 1)
3(2 + 4) - (3 - 1)
Now, simplify both parentheses:
3(6) - 2
18 - 2
Finally, simplify:
16
Answer:
Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old
Step-by-step explanation:
Data given
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
And replacing we got:
Now we can calculate the critical value but first we need to find the degreed of freedom:
So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:
Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old
