a car dealer buys a luxury car for $65,200 and sells it for 85,650. What is the percent increase in the price of the cara car de
1 answer:
Answer:
The percent increase in the price of the car is <u>31.36%</u>.
Step-by-step explanation:
Given:
A car dealer buys a luxury car for $65,200 and sells it for $85,650.
Now, to find the percent increase in the price of the car.
Cost price of the luxury car = $65,200.
Sell price of the luxury car = $85,650.
So, the price of the car increased by:
Sell price of the luxury car - Cost price of the luxury car.
<em>Thus, the price of car increased by $20,450.</em>
Now,to get the percent increase in the price of the car:
Therefore, the percent increase in the price of the car is 31.36%.
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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.
Answer:
1.0 mL each day
Step-by-step explanation:
The required daily dosage of vitamin D is 10 micrograms.
The supplement contains 5 micrograms per 0.5mL.
The volume of supplement necessary to fulfill the daily requirements of vitamin D is:
She should administer 1.0 mL of the supplement each day.