Answer:
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For the population, we have that:
Mean = 15
Standard deviaiton = 12
Sample of 30
By the Central Limit Theorem
Mean 15
Standard deviation 
Approximately normal
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.
Answer:
Step-by-step explanation:
To find the diagonal, use Pythagorean theorem,
diagonal² = side² + side²
= (9√3)² + (9√3)²
= 9²(√3)² + 9²(√3)²
= 81*3 + 81*3
= 243 + 243
= 486
diagonal = √486 = 22.05 units
So your present value is 7750. this is what you have now. they give you the intesrest rate which is 4% and how the amount of year is 5.
recall the formula for a compounded continously is
P×e^(r×t)
the answer is 7750×e^(.04×5)
Answer:
21/676
Step-by-step explanation:
The total number of letters of the alphabet will be the total outcome which is 26 alphabets.
If consonant numbers are drawn out, the probability of drawing a consonant will be total consonants/total alphabets
Total consonants in the alphabet is 21.
Probability of drawing consonants = 21/26.
Since all the consonants are replaced before drawing a Z, our total outcome will not change i.e it will still be 26.
Probability of drawing a letter Z will be 1/26.
Therefore, the probability of drawing a consonant,replacing it, and then drawing a Z will be;
21/26×1/26
= 21/676
Answer:
1-8
Step-by-step explanation:
I was only able to finish one through eight The rest follow the same concept though