Answer:
1. a) (1-1) (1-2) (1-3) (2-1) (2-2) (3-1)
2. a) (5-1) (5-2) (5-3) (5-4) (5-5) (5-6)
3. d) (1-3) (2-3) (3-1) (3-2) (3-4) (3-5) (3-6) (4-3) (5-3) (6-3)
Step-by-step explanation:
1. added the numbers in the parenthesis and A had all of the sums less than 5
2. A had all the 5s in the first spot
3. D had the 3s in each parenthesis
If you would like to find the decimal number which is equivalent to 1/10, you can do this using the following step:
1/10 = 0.1
The correct result would be B. 0.1.
Answer:
z= 3.63
z for significance level = 0.05 is ± 1.645
Step-by-step explanation:
Here p = 42% = 0.42
n= 500
We formulate our null and alternative hypotheses as
H0: p= 0.42 against Ha : p> 0.42 One tailed test
From this we can find q which is equal to 1-p= 1-0.42 = 0.58
Taking p`= 0.5
Now using the z test
z= p`- p/ √p(1-p)/n
Putting the values
z= 0.5- 0.42/ √0.42*0.58/500
z= 0.5- 0.42/ 0.0220
z= 3.63
For one tailed test the value of z for significance level = 0.05 is ± 1.645
Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01

X[bar] ± 
174.5 ± 
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!
The answer is B.Center (2 -8) Radius =4