Since we are not given any information about the proportion, we will assume the sample proportion to be 0.50
so,
p = 0.50
The Error is 10% percentage point. This means that on either side of the population proportion the error is 5% so E = 0.05
z = 1.645 (Z value for 90% confidence interval)
The margin of error for population proportion is calculated as:
This means 271 students should be included in the sample
I am pretty sure your answer is going to be B, <span>There are two outliers, indicating very few books were rented out on those two days. I hope this helps!! :)</span>
Answer:
A=8.4063
Step-by-step explanation:
Be the functions:
according the graph:
=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B3%7D%7Bx%7D%20%7D%20%5C%2C%20dx%20-%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B3%7D%7Bx%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%20%3D3%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B1%7D%7Bx%7D%20%7D%20%5C%2C%20dx%20-3%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%3D3%28%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B1%7D%7Bx%7D%20%7D%20%5C%2C%20dx%20-%5Cint%5Climits%5E1_7%20%7B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%29%3D3%5Blnx-%5Cfrac%7B1%7D%7Bx%7D%5D%281-7%29%3D3%5B%28ln7-ln1%29-%28%5Cfrac%7B1%7D%7B7%7D-1%29%5D%3D3%5B%281.945-0%29-%280.1428-1%29%5D%3D3%2A%281.945%2B0.8571%29%3D3%2A2.8021%3D8.4063u%5E%7B2%7D)
One of the easiest question I've seen.
It's a scale factor of 5 because,
5*5=25
5*3=15
5*4=20
Hope this helps!
Answer:
a) 0.88
b) 0.35
c) 0.0144
d) 0.2084
e) 0.7916
Step-by-step explanation:
a) The probability of a peanut being brown is 12/100 = 0.12. Hence the probability of it not being brown is 1-0.12 = 0.88
b) 12% of peanuts are brown, 23% are blue. So 35% are either blue or brown. The probability of a peanut being blue or brown is, therefore 35/100 = 0.35.
c) 12% of peanuts are red, so the probability of a peanut being red is 12/100 = 0.12. In order to calculate the probability of 2 peanuts being both red, we can assume that the proportion doesnt change dramatically after removing one peanut (because the number of peanuts is absurdly high. We can assume that we are replenishing the peanuts). To calculate the probability of 2 peanuts being both red, we need to power 0.12 by 2, hence the probability is 0.12² = 0.0144.
d) Again, we will assume that the probability doesnt change, because we replenish. The probability of a peanut being blue is 0.23. The probability of it not being blue is 0.77, so the probability of 6 peanuts not being blue is obtained from powering 0.77 by 6, hence it is 0.77⁶ = 0.2084
e) The event 'at least one peanut is blue' is te complementary event of 'none peanuts are blue', so the probability of this event is 1- 0.2084 = 0.7916
