Answer:
Step-by-step explanation:
The urban planner collects travel times from a random sample of 125 commuters in the San Francisco Bay Area. A traffic Study from last year claimed that the average commute time in the San Francisco Bay Area is 45 min. The urban planner will see if there is evidence the average commute time is greater than 45 minutes
( Here in this case, Null hypothesis will be Η :μ = 45
And the Alternate Hypoyhesis will be H, :μ> 45 )
C. The urban planner asks a random Sample of 100 commuters in the San Francisco Bay Area to record travel times on a Tuesday morning. One year later, the urban planner asks the same 100 commuters to record travel times on a tuesday morning . The urban planner will see the difference in commute time shows an increase.
Here in this case the null hypothesis will be, H₀ : 
= 0
And the Alternate Hypothesis will be H, : <0 The commute time after 1 year is more
Answer:
Step-by-step explanation:
y = 50x - 165
Answer:
They would by able to go 4 miles.
Step-by-step explanation:
Point B is located at point (4, -2)
Answer
a. 28˚
b. 76˚
c. 104˚
d. 56˚
Step-by-step explanation
Given,
∠BCE=28° ∠ACD=31° & line AB=AC .
According To the Question,
- a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°
- b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚
thus , ∠ABC=76° .
- c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,
Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°
Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)
∠DAB = 180° - 107° ⇔ 73°
& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°
Now, In Triangle ADC Sum of angles in a triangle is 180°
∠ADC = 180° - (31° + 45°) ⇔ 104˚
- d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre
For Diagram, Please Find in Attachment
