Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation,
Z value for 108,
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Answer and Explanation:
Gas chromatography separates compounds depending on their **polarity and volatility**. Benzene, m-xylene, and toluene have similar **polarities**, therefore, the main basis for separation is **volatility**. The more volatile a component the ** higher its vapor pressure**, hence the more time it spends in the **gaseous mobile phase**, giving it a **shorter** retention time. Therefore, components of a liquid mixture will elute in order of **increasing boiling points/decreasing volatilities/increasing polarities with the stationary phase**.
Answer:
Velocity of ball B after impact is
and ball A is 
Explanation:
= Initial velocity of ball A

= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have

Coefficient of restitution is given by



Adding the above two equations we get



From the conservation of momentum along the plane of contact we have


Velocity of ball B after impact is
and ball A is
.
Answer:D. Gunter's Chain
Explanation:I know this because a gunter's chain is used for plots of land to be accurately surveyed and plotted, for legal and commercial purposes.
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds