Answer:
0.5%/year
24.2%
Step-by-step explanation:
Estimate the average yearly increase in the percentage of first-year college females claiming no religious affiliation
Percentage of females by year:
1980 = 6.2%
1990 = 10.8%
2000 = 13.6%
2012 = 21.7%
Average yearly increase :
Percentage increase between 1980 - 2012 :
2012% - 1980% = ( 21.7% - 6.2%) = 15.5% increase over [(2012 - 1980)] = 32 years
15.5 % / 32 years = 0.484375% / year = 0.5%/year
b. Estimate the percentage of first-year college females who will claim no religious affiliation in 2030,
Given an average increase of 0.484375% / year
(2030 - 1980) = 50 years
Hence by 2030 ; ( 50 years × 0.484375%/year) = 24.218% will claim no religious affiliation.
=24.2% (nearest tenth)
m□ebd=4 x-8 and m□ebc=5 x+20
This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.
∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.
i.e
∠ ebd + ∠ebc = 180°
4 x- 8 + 5x + 20= 180°
adding like terms
⇒ 9 x +12 =180°
⇒ 9 x = 180° - 12
⇒ 9 x = 168°
⇒ x =( 168/9)°=(56/3)°
now m□ebc =5 x +20
= 5 × 56/3 + 20
= 280/3 + 20
=340/3
m□ebc=( 340/3)°
So, solution set is x =(56/3)° and m□ebc =(340/3)°
40 divided by 8 is equal to 5 so 5 is going to be the zoom size
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.