Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
Answer: -0.729
Step-by-step explanation:
Answer: 32.8kgm^2/s
Step-by-step explanation:
1. Angular momentum is a product of linear momentum and the distance of the object from a rotation axis.
2. Linear momentum (L) is given as :
L = mv
Where: m= mass = 4kg
v= velocity= 2m/s
=> L = 8kgm/s
Angular momentum (A) = L x centre distance (d) of object from a rotation axis
d = √(0.9-0)^2 + (10-6)^2 = 4.1m
=> A = 8 x 4.1 = 32.8kgm^2/s.
Answer:
-31
Step-by-step explanation:
If we continue to subtract 5 from the previous term, the sequence continues ...
4, -1, -6, -11, -16, -21, -26, -31, -36, ...
The 8th term is -31.
_____
This is an arithmetic sequence with first term a1 = 4 and common difference d = -5. The general term (an) of such a sequence is ...
an = a1 +d(n -1)
Filling in the values for this sequence, the general term is ...
an = 4 -5(n -1)
Then the 8th term is ...
a8 = 4 -5(8 -1) = 4 -35
a8 = -31
