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.
You know a1.
So find a2, a3, and so on until a7.
a(1) = 12
a(2) = 16
a(3) = 20
a(4) = 24
a(5) = 28
a(6) = 32
a(7) = 36
Each is 4 more than the previous.
Answer: Length = 28 feet and width = 9 feet
Step-by-step explanation:
Let x = width of the wall.
Then, Length of the wall = 10+2x
Since , Area of rectangle = length x width
Then, as per given,
Since width cannot be negative , so x = 9
So width = 9 feet and length = 10+2(9)=10+18 = 28 feet
Hence, length is 28 feet and width is 9 feet of the wall of the barn.
Answer:
The degrees of freedom for this sample are 27.
The sample size to get a margin of error equal or less than 0.3656 is n=4450.
Step-by-step explanation:
The degrees of freedom for calculating the value of t are:
With 27 degrees of freedom and 95% confidence level, from a table we can get that the t-value is t=2.052.
The sample size to get a margin of error equal or less than 0.3656 can be calculated as:
Answer:
it's the first one because it is split as the paragraph said it should
