Answer:
A) The mean of the chi-square distribution is 0
A) is not a property of chi square distribution.
Step-by-step explanation:
We have to find the properties of a chi square test.
A) False
The mean of a chi square distribution is equal to the degree of freedom.
B) True
The chi-square distribution is non symmetric.
C) True
The chi square value can be zero and positive.
It can never be negative because it is based on a sum of squared differences .
D) True
The chi-square distribution is different for each number of degrees of freedom.
When we are working with a single population variance, the degree of freedom is n - 1.
Answer:The amount of paint that was sold altogether is 173.36 litres
Step-by-step explanation:
The total amount of paint that the paint shop stocks is 1800 litres.
24% of the paint is white. It means that the amount of white paint would be
24/100 × 1800 = 0.24 × 1800 = 432 litres.
The amount of the remaining paint other than white would be
1800 - 432 = 1368 litres
The shops sells 18% of the white paint. This means that the amount of white paint sold by the shop will be
18/100 × 432 = 0.18 × 432 = 77.6 litres.
The shops sells 7% of the rest of the paint.
This means that the amount of the rest paint sold by the shop will be
7/100 × 1368 = 0.07 × 1368 = 95.76 litres.
The amount of paint that was sold altogether would be
77.6 + 95.76 = 173.36 litres
Answer: He cannot see a ship that is 4.5 miles away.
Step-by-step explanation:
Since we have given that
Here, h is the eye level above the water
d is the distance away.
If h = 12 feet
We will get :
Hence, it is 10.39 feet away when eye level is 12 feet above the water.
Therefore , he cannot see a ship that is 4.5 miles away.
Answer:
Neither are correct.
Step-by-step explanation:
The number of legs = the number of ants * 6.
The relationship is L = 6 a so neither are correct.
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
