Answer:
x = 47°
Step-by-step explanation:
The secant- secant angle (x) is half the difference of the intercepted arcs, that is
x = (133 - 39)° = 0.5 × 94° = 47°
To decrease an amount by 60% what single multiplier would you use
1 answer:
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This question is incomplete, the complete question is;
The following is based on information from The Wolf in the Southwest: The making of an Endangered Species by David E. Brown ( University of Arizona Press). Before 1918the proportion of female wolves in the general population of all southwestern wolves was about 50%. However, after 1918, southwestern cattle ranchers began a widespread effort to destroy wolves. In a recent sample of 34 wolves, there were only 10 females. One theory is that male wolves tend to return sooner than females to their old territories where their predecessors were exterminated. Do these data indicate that the population proportion of female wolves is now less than 50% in the region? Use ∝ = 0.01
Answer:
Since p-value (0.0082) < ∝ ( 0.01 ), we reject null hypothesis.
Therefore, at 0.01 significance level, we have sufficient evidence that the population proportion of female wolves is now less than ( 0.5 ) 50% in the region.
Step-by-step explanation:
Given the data in the question;
Hypothesis;
H₀ : p ≥ 0.5 { 50% }
H₁ : p < 0.5
This is a lower tailed test { H₁ : p < 0.5 }
sample size n = 34
number of female wolves x = 10
sample proportion p" = x / n = 10 / 34 = 0.2941176
claimed proportion P = 0.5
significance level ∝ = 0.01
we determine the standard deviation of p"
σ = √[ (p(1 - p)) / n ] = √[ (0.5(1 - 0.5)) / 34 ]
= √[ (0.5 ×0.5) / 34 ]
σ = 0.08575
Test statistics
= (p" - 0.5) / σ
= (0.2941176 - 0.5) / 0.08575
= -2.40
so
Test statistic : -2.40
Since its a lower tailed test;
P-value = P( Z < ) = P( Z < -2.40 ) = 0.0082
Rejection criteria: Reject H₀ if p-value < ∝
Decision:
Since p-value (0.0082) < ∝ ( 0.01 ), we reject null hypothesis.
Therefore, at 0.01 significance level, we have sufficient evidence that the population proportion of of female wolves is now less than ( 0.5 ) 50% in the region.