Answer:
5.26%
Step-by-step explanation:
15/285=x/100. 15 is the amount of weight he wants to lose. x=5.26%
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:
the answer is $40.5
Step-by-step explanation:
the basic formula of finding our interest is I = PxRxT
P (principal) = 675
R (rate) = 10% (we have to turn it into a decimal which is 0.010)
T (time) = 6 years
so :
I = 675 x 0.010 x 6
I = 40.5
They are inverses if f( (g(x)) = x
f((g(x)) = (1/2) (2x + 4) - 2 = x + 2 - 2 = x
so this proves that g(x) is the inverse of f(x)
the other inverse g((f(x)) should also be = x
