It rounded to the nearest dime what is the greatest amount of money that rounds to $105.40

To achieve the desired output, first use the machine with the function y = x^2 - 6, followed by the machine that computes y = sqrt(x-5). This way, when you input 6, the output from the first machine is calculated as x = 6, yielding y = 6^2 - 6, resulting in 30 as the input for the second machine. The second machine then processes this to provide the final output of sqrt(30 - 5), which equals sqrt(25) = 5. Alternatively, to obtain a negative final output, you should first utilize the machine with the function y = sqrt(x-5). Assuming you start with the value x = 9, the first machine computes this to sqrt(9-5), which is sqrt(4) = 2. Then, the second machine converts y to 2^2 - 6, leading to a result of 4 - 6 = -2.

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.

<span>The accurate answer is "<span>The experimental probability is 1/15 lower than the theoretical probability."

The attached table illustrated in the image indicates that the theoretical probabilities are: 2/5 for drawing a Jack, 4/15 for a Queen, and 1/3 for a King, based on the frequency of each card in the set.
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Answer:

The answer is

Step-by-step explanation:

After operating under these conditions for a period of 6 months, he selects a simple random sample of 15 days. He logs the hours he walked each day as tracked by his fitness monitor, alongside his recorded billable hours from a work application on his computer.

The calculated slope is -0.245

The number of samples is n = 15

The standard error is 0.205

The confidence level is 95

The significance level is (100 - 95)% = 0.05

The degrees of freedom is n - 2 = 15 - 2 = 13

Critical Value = 2.16 = [using excel = TINV (0.05, 13)]

Marginal Error = Critical Value * standard error

= 2.16 * 0.205

= 0.4428

Let x represent the number of apples and y the number of oranges.

Starting with the first scenario, since each apple is priced at $0.24 and each orange at $0.80, we have:
total spent on apples = 0.24x and total spent on oranges = 0.8y
Given that the total expenditure is $12, the first equation is:
0.24x + 0.8y = 12

For the second scenario, the total number of pieces of fruit purchased is 20, thus the second equation is:
x + y = 20

By graphing these equations, one can identify a potential combination at the intersection of the two lines.