a.When the numerator and the denominator have no common factor other than 1; he fraction is either a proper fraction or mixed nu
1 answer:
Answer:
The fraction is a proper fraction (and irreducible)
Step-by-step explanation:
Given
Required
The type of fraction
A fraction with the given features is a proper fraction and cannot be further reduced.
Take for instance:
The only common factor between is 1
See that 2/7 is a proper fraction and also 2/7 cannot be reduced (simplified)
<em>The above logic is applicable to all fraction that have the attribute stated in the question (e.g. 3/7, 3/8, 4/9, 7/10, etc.)</em>
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Answer:
(a) The median is 1478
(b) Null hypothesis H₀ : μ₁ = μ₂
Alternative hypothesis Hₐ : μ₁ > μ₂
(c) The test statistic is 6.678155
(d) The p value is 1.2×10⁻¹¹
(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high
Step-by-step explanation:
(a) Here we have the data as follows;
1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377
The median is = 1478
Therefore we have above the median
1623 1586 1574 1554 1544 1531 1519 1513 1491 1484
The mean, = 1541.9
Standard deviation, σ₁ = 41.38224257
n₁ = 10
Below the median
1471 1458 1456 1458 1453 1438 1428 1405 1395 1377
The mean, = 1433.9
Standard deviation, σ₂ = 30.04812806
n₂ = 10
(b) Null hypothesis H₀ : μ₁ = μ₂
Alternative hypothesis Hₐ : μ₁ > μ₂
(c) The formula for t test is given by;
df = 10 - 1 = 9, α = 0.05
Therefore, the test statistic = 6.678155
(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹
Critical z at 5% confidence level = 1.645
Since P << 0.05, e reject the
(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.