Describe the operations used to solve the equation. 5
2
3. X-2 X+1
X-2 = 6
5x - 12 = -4x + 6
+4x
+4x
9x - 12 =
6
+12
+12
9x = 18
9x
9
18
9
Х
= 2

Acaterer knows he will need 60, 50, 80, 40 and 50 dinner napkins on five successive evenings. he can purchase new napkins initially at 25 cents each, after which he can have dirty napkins laundered by a fast one-day laundry service (i.e., dirty napkins given at the end of the day will be ready for use the following day) at 15 cents each, or by a slow two-day service at 8 cents each or both. the caterer wants to know how many napkins he should purchase initially and how many dirty napkins should be laundered by fast and slow service on each of the days in order to minimize his total costs. formulate the caterer’s problem as a linear program as follows (you must state any assumptions you make): a. define all variables clearly. how many are there? b. write out the constraints that must be satisfied, briefly explaining each. (do not simplify.) write out the objective function to be minimized. (do not simplify.)

According to a research study, parents with young children slept 6.4 hours each night last year, on average. a random sample of 18 parents with young children was surveyed and the mean amount of time per night each parent slept was 6.8. this data has a sample standard deviation of 0.9. (assume that the scores are normally distributed.) researchers conduct a one-mean hypothesis at the 5% significance level, to test if the mean amount of time parents with young children sleep per night is greater than the mean amount of time last year. the null and alternative hypotheses are h0: μ=6.4 and ha: μ> 6.4, which is a right-tailed test. the test statistic is determined to be t0=1.89 using the partial t-table below, determine the critical value(s). if there is only one critical value, leave the second answer box blank.

(co 3) the times that customers spend in a book store are normally distributed with a mean of 39.5 minutes and a standard deviation of 9.4 minutes. a random sample of 25 customers has a mean of 36.1 minutes or less. would this outcome be considered unusual, so that the store should reconsider its displays? no, the probability of this outcome at 0.035, would be considered usual, so there is no problem yes, the probability of this outcome at 0.035, would be considered unusual, so the display should be redone no the probability of this outcome at 0.359 would be considered usual, so there is no problem yes, the probability of this outcome at 0.965 would be considered unusual, so the display should be redone