You ask 40 students which of three items from the cafeteria they like the best. You record the results on the piece of paper sho

Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

The determined Z value is 2.53

In this case, the critical value Z is 1.96 at the 0.05 significance level

Step(iii):-

Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

We accept the alternative hypothesis

Conclusion:-

We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds

Answer:

link:

Step-by-step explanation:

Explanation:

Step-by-step clarification:

Referring to step 6

(b² — 4ac) / 4a² = (x + b/2a)²

The mistake in the question is that it should be (x + b/2a)²

According to step 7

±√(b² —4ac) /2a = x + b/2a

The error in the question is that it should be divided by 2a, not 1a.

1. The transition from step 6 to step 7 involves taking the square roots of both sides

(b² — 4ac) / 4a² = (x + b/2a)²

Taking the square of both sides

√(b²—4ac) / √4a² = √(x + b/2a)²

√(b²—4ac) / 2a = x + b/2a

This forms step 7 correctly.

Next, subtracting b/2a from both sides

√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a

√(b²—4ac) / 2a — b/2a = x

(√(b²—4ac)  — b)/2a = x

x = [—b ± √(b²—4ac)] / 2a

This gives the desired formula.

The discriminant is D = b²—4ac.

To formulate the system, it's necessary to consider the slope of each line along with at least one point from each line. The two lines will connect each plane's location to their destination airport. It's important to note that the airport's coordinates represent the intersection of these two lines, corresponding to the solution of the system. First, the slope of the line from airplane one to the airport is: m = 2; this can be observed by plotting the two points. From airplane 1's location, the rise is 8 units while the run is 4 units to reach the airport, making the slope 8 divided by 4 = 2. We then insert the slope and point (2,4) into the point-slope form: y - 4 = 2(x - 4), which can be rearranged to standard form 2x - y = 0. For airplane two, the slope to the airport is obtained by observing the vertical decrease of 3 and a horizontal increase of 9 as we move from the airport to airplane 2. We then substitute the slope and the point (15,9) into the point-slope form: y - 9 = -1/3(x - 15), which can be rearranged to the standard form: x + 3y = 42. Consequently, the system of equations is: 2x - y = 0 and x + 3y = 42. Multiplying the first equation by 3 produces a system of: 6x - 3y = 0 and x + 3y = 42. Adding these equations results in the equation 7x = 42. Thus, x = 6, and by substituting this value back into 2x - y = 0, we determine y = 12. Thus, we demonstrate that the airport's coordinates do indeed comprise the solution to our system.