Hello there,
best way to do this one is set up your story line....
4h + 8h + 50 = 164$...
let's make this easier to look at and combine like terms..
12h + 50 = 164$...
Beautiful!!! Now let's try and get the h alone?...
minus 50 to both sides...
12h = 114$..
now divide both sides by 12...
H= 9.50$/hour
Looks like you made 9.50$ an hour but Hey, at least you got that bonus ;)
Answer: 0.02
Step-by-step explanation:
OpenStudy (judygreeneyes):
Hi - If you are working on this kind of problem, you probably know the formula for the probability of a union of two events. Let's call working part time Event A, and let's call working 5 days a week Event B. Let's look at the information we are given. We are told that 14 people work part time, so that is P(A) = 14/100 - 0.14 . We are told that 80 employees work 5 days a week, so P(B) = 80/100 = .80 . We are given the union (there are 92 employees who work either one or the other), which is the union, P(A U B) = 92/100 = .92 .. The question is asking for the probability of someone working both part time and fll time, which is the intersection of events A and B, or P(A and B). If you recall the formula for the probability of the union, it is
P(A U B) = P(A) +P(B) - P(A and B).
The problem has given us each of these pieces except the intersection, so we can solve for it,
If you plug in P(A U B) = 0.92 and P(A) = 0.14, and P(B) = 0.80, you can solve for P(A and B), which will give you the answer.
I hope this helps you.
Credit: https://questioncove.com/updates/5734d282e4b06d54e1496ac8
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Answer : a link for your answer<em> https://quizlet.com/140197143/surface-area-and-volume-surface-area-of-rectangular-pyramids-flash-cards/</em>
<em>Step-by-step explanation:</em>
