Jessie thinks that the first step in solving the equation shown below could be to use the addition property of his equality. His

Hi! The goal of the Chi-Square Goodness of Fit test is to determine if observed frequencies of a categorical variable align with the expected historical or theoretical values in the population. Having the sales proportions of the top-five compact cars, we compare them against 400 compact car sales data from Chicago to see if there are discrepancies. Specifically, we have: - Chevy Cruze 24% ⟹ P(CC) = 0.24 - Ford Focus 21% ⟹ P(FF) = 0.21 - Hyundai Elantra 20% ⟹ P(HE) = 0.20 - Honda Civic 18% ⟹ P(HC) = 0.18 - Toyota Corolla 17% ⟹ P(TC) = 0.17 The hypotheses established are: H₀: P(CC) = 0.24; P(FF) = 0.21; P(HE) = 0.20; P(HC) = 0.18; P(TC) = 0.17 H₁: There is a discrepancy between expected and observed outcomes. With α set at 0.05, the statistic calculated is based on Oi (observed frequency) and Ei (expected frequency). The initial step involves calculating expected frequencies using: Ei = n * Pi, where Pi is the theoretical proportion for each category stated in the null hypothesis. The test conducted is right-tailed, and so is the p-value, calculated as: P(X²₄ ≥ 11.23) = 1 - P(X²₄ < 11.23) = 1 - 0.98 = 0.02. Since the p-value is lower than α, we reject the null hypothesis, indicating that Chicago's market shares for the five compact cars differ from those reported by Motor Trend.

Answer:

The first equation is y = 1.5x - 6

The second equation is y = -4x + 6.1

The solution is (2.2, -2.7)

Step-by-step explanation:

Response:

$54

Detailed steps to explanation:

Initially

45 multiplied by.05 equals 2.25

Next

45 added to 2.25

Following that

45 multiplied by.15 equals 6.75

Finally

45 plus 2.25 plus 6.75

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

The function

represents a parabola positioned at the vertex (5, 3).

Attached here is the graph depicting this function.