Find the value of the linear correlation coefficient r. The paired data below consist of the temperatures on randomly chosen day

a) P(identified as explosive) equals P(actual explosive & identified as explosive) + P(not explosive & identified as explosive) = (10/(4*10^6))*0.95+(1-10/(4*10^6))*0.005 = 0.005002363. Thus, the probability that it actually contains explosives given that it's identified as containing explosives is (10/(4*10^6))*0.95/0.005002363 = 0.000475. b) Let the probability of correctly identifying a bag without explosives be a. Therefore, a = 0.99999763, approximately 99.999763%. c) No, even if this becomes 1, the true proportion of explosives will always be below half of the total detected.

Answer:

There are 369 students who enrolled in either calculus or discrete mathematics.

Step-by-step explanation:

I'll create a Venn diagram to illustrate these figures.

Let’s define:

A represents the count of students who completed calculus.

B stands for those who took discrete mathematics.

We have the following:

Here, a denotes the students who studied calculus exclusively, while represents those who took both subjects.

Using the same reasoning:

There are 188 students taking both calculus and discrete mathematics.

This implies that

A total of 212 students studied discrete mathematics.

<pBased on this,

345 students have attended a calculus course.

<pSo, from this information,

We find the total number of students who took either calculus or discrete mathematics.

A total of 369 students participated in either calculus or discrete mathematics.

Answer:

Hello! Your question appears to be missing details; here is the complete version

Ix = 0   Ux =

Iz = 0   Uz =

Iy = 5   Uy = 10

Step-by-step explanation:

Ix = 0   Ux =

Iz = 0   Uz =

Iy = 5   Uy = 10

This provides a comprehensive solution below