Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

The function's constant term is 5. Step-by-step explanation: we have where b represents the y-intercept or the constant term of the function. The x-intercept indicates the value of x when the function equals zero. Thus, for x = -3, f(x) = 0 and for x = -5, f(x) = 0. Substitute either of the intercepts into the function. Check the other intercept as well. For x = -5, this is proven to be true. Hence, the consistent term in the function is 5.

Answer:

Step-by-step explanation:

The goal here is to identify whether the provided data set adheres to a normal distribution, and consequently, if a t-test is suitable for application.

To address this, we will construct a box plot using MINITAB software

The guidelines for this are as follows:

Input your data into the worksheet

Then navigate to Graph→ Boxplot→ EDA→ Boxplot

Under One Y, choose simple, and click OK

In Variables, enter measurements. Then click OK

Thus, for the first dataset (a) 213.03 212.95 213.04 213.00 212.99 213.01 221.03 213.05 as illustrated in the attached file in word document for the boxplot below;

We would observe an outlier in the data set which suggests that it does not conform to the normal distribution assumption; hence, the use of the t-distribution is not advised.

(b) Referring to the attached document below; we will notice that the boxplot does not present any outliers. Furthermore, it is evident that the distribution is right skewed. Thus, this dataset fails the necessary assumptions, making the t-distribution inappropriate for this data.

(c)

In the document provided below; we will find the boxplot without any outliers. Additionally, it is clear that the distribution exhibits a left skew. Consequently, the data does not meet the required assumptions, resulting in the t-distribution being unsuitable for this dataset.

Try this approach (refer to the attachment, which outlines three steps).

Having no profit indicates that P(x)=0: x³-4x²+3x-12=0
factor this by regrouping: x²(x-4)+3(x-4)=0
extract (x-4): (x-4)(x²+3)=0
thus, x=4 or x²=-3
since x² cannot be negative, the only viable solution is x=4.