The price of each ticket will be the slope found from the data.
Slope=m=(y2-y1)/(x2-x1)=(177.5-146.25)/(18-13)
m=6.25 (so the cost of the tickets is $6.25)
Now we can use any data point to solve for b, or the y-intercept of the line:
y=6.25x+b, using (177.5, 18)
177.5=6.25(18)+b
b=$65.00d
So he started with $65.00
Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
First you need to know the definition of a complementary angle. That means that two angles add up to 90 degrees, or a right angle.
To find an angle complementary to a 78 degree angle, you need to subtract 78 from 90.
90 - 78 = 12
Only the third one makes x=-1 true
Answer:
y = (x/(1-x))√(1-x²)
Step-by-step explanation:
The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:
x = r·cos(θ)
y = r·sin(θ)
x² +y² = r²
__
r = sec(θ) -2cos(θ)
r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)
r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²
(x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations
x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor
y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)
__
The attached graph shows the equivalence of the polar and rectangular forms.
