Answer:
A)n= 703.96
B)n= 602.308
Step by step Explanation:
Given that you want to be 99% confident that the sample percentage is within 3.1 percentage points of the true population percentage.
Then z/2 = 1.645
And M = 3.1% = 0.031
A)Nothing is known therefore,
p = q = 0.50
E=0.031
For 90% confidence, z = 1.645
n = (zα/₂)²(p)(1-p)/M²
n= 1.645²× 0.5 × 0.5/0.031²
n= 703.96
Therefore, 703.96randomly selected air passengers must be surveyed to be 99%
B)we know that recent surveys surgest that about 38% of all air passengers prefer an aisle seat, thus p = 35% = 0.35
n = (zα/₂)²(p)(1-p)/M²
n= (1.645²× 0.31 × 0.69)/0.031²
n= 602.308
Hence, 602.308 randomly selected air passengers must be surveyed to be 90% confident that the sample percentage is within 3.1 percentage points of the true population percentage.
Y=2 yes (the slope is 0 and the y intercept is 2)
Answer:
the answer is 55038271829339
Answer:
Green and Blue ribbon
Step-by-step explanation:
Given
They collect
Red Ribbon = ½ mile
Green Ribbon = ⅛ mile
Blue Ribbon = ¼ mile
To find?
Which colors of ribbons will be collected at the ¾ mile mark.
The interpretation of the question is to test which of the above fractions can divide ¾ without a remainder; in other words, multiples of ¾.
Testing each fraction.
Red Ribbon = ½
¾ ÷ ½
= ¾ * 2
= 3/2
= 1.5 or 1 Remainder 1
This is not an exact multiple of ¾. So, the red ribbon won't be passed here.
Green Ribbon = ⅛
¾ ÷ ⅛
= ¾ * 8
= 24/8
= 3
This is an exact multiple of ¾. So, the green ribbon will be collected.
Testing the last ribbon
Blue = ¼
¾ ÷ ¼
= ¾ * 4
= 3
This is an exact multiple of ¾. So, the blue ribbon will be collected.
Hence, the green and blue ribbons will be collected at ¾ mile mark
I think it would take about 1 hour and 17 minutes
