What is the domain of the function
1 answer:
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Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
Answer:
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Step-by-step explanation:
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.
At the null hypothesis, we test if the mean cost is still the same, that is:
At the alternative hypothesis, we test if the mean cost has increased, that is:
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question
In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
13252 is tested at the null hypothesis:
This means that
The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.