Answer:
$120.
Step-by-step breakdown:
Libby sold $60 worth of vegetables at the farmer's market.
Jess sold 3/4 of what Libby sold, so let's calculate that: 3/4 of 60.
Thus, Jess made $45.
Now, Stephanie sold one-third of Jess's total, necessitating the calculation of one-third of 45.
This means Stephanie sold $15 worth of vegetables.
Putting together Libby's, Jess's, and Stephanie's sales, we arrive at the collective total earned by all three.
Consequently, their total sales amount to $120.
You can calculate 400+200 or sum 368+231= 599 and then round your result, which gives 600
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.
We aim to verify the assertion that generally, 10% of students repeat a course, leading us to this hypothesis setup:
Null hypothesis:
Alternative hypothesis.
The most fitting choice for this scenario is:
d) H0:p=0.1 vs. H1:p ≠ 0.1.
For this case, the provided information includes: the number of students repeating the course, the selected sample size, and the estimated proportion of repeaters. We are testing the claim that generally, 10% of students retake classes, which will be validated through established hypotheses.
