A restaurant owner plans to use x tables seating 4, y tables seating 6, and z tables seating 8, for a total of 20 tables. When f

The total number of juices equals 27, with the probabilities for each type being as follows: apple juice = 12/27, grape juice = 15/27, sugar-free = 14/27, and not sugar-free = 13/27. Since it has already been established that the chosen juice is not sugar-free, we do not need to factor that probability into our calculations. Of the apple juices, 9 are sugar-free, leaving 3 that are not, and for the grape juices, 5 are sugar-free, resulting in 10 that are not. Consequently, among 13 juices that are not sugar-free, 10 are grape juice, so the likelihood of selecting a non-sugar-free grape juice is 10/13. Therefore, the answer is A). Sorry if my explanation was lengthy; I tend to elaborate.

Answer:

Step-by-step explanation:

According to the attached figure, Jaxon is positioned 10 blocks east and 5 blocks north.

Thus, Jaxon's x-coordinates are 10 units, while the y-coordinates are 5 units.

Therefore, the coordinates are given by (10,5)

Similarly, Isaac's coordinates are determined based on his distances towards the west and south.

Hence, the x-coordinates equal 8 units and the y-coordinates equal 15 units.

Thus, Isaac’s coordinates are depicted as (-8,-15).

The response indicates that the test comprises 10 questions worth 3 points each and 14 questions worth 5 points. If there are any queries, feel free to ask!!! Thank you!

Given:
1 pack = 5 pencils and cardboard.
1 pack should weigh between 60 grams and 95 grams

60g < x < 95g; where x signifies 1 pack.

Cardboard: 15 grams.

95g - 15g = 80g represents the maximum total weight of 5 pencils.
80g / 5 = 16g is the maximum weight for a single pencil.

60g - 15g = 45g is the minimum total weight of 5 pencils.
45g / 5 = 9g is the minimum weight for a single pencil.

9 < x < 16; where x represents a single pencil in the pack.

Response:

Detailed explanation:

For private institutions,

n = 20

Average, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(sum of (x - mean)²/n

Sum of (x - mean)² = (43120 - 34623.05)^2 + (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public institutions,

n = 20

Average, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Sum of (x - mean)² = (25469 - 25063.15)^2 + (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This involves two independent samples. Define μ1 as the mean out-of-state tuition for private institutions and μ2 as the mean out-of-state tuition for public institutions.

The random variable represents μ1 - μ2 = the difference between the mean out-of-state tuition for private vs. public institutions.

The hypothesis is established as follows. The correct choice is

-B. H0: μ1 = μ2; H1: μ1 > μ2

As the sample standard deviation is known, the test statistic is calculated using the t test formula:

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The method for finding degrees of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

The probability value is obtained from the t test calculator. It is

p value = 0.000065

Given that alpha, 0.01 > the p value, 0.000065, we will reject the null hypothesis. Hence, at a significance level of 1%, the mean out-of-state tuition for private institutions is statistically significantly greater than that of public institutions.