Answer:
arithemetic sequence i guess
Answer: 0.1457
Step-by-step explanation:
Let p be the population proportion.
Given: The proportion of Americans who are afraid to fly is 0.10.
i.e. p= 0.10
Sample size : n= 1100
Sample proportion of Americans who are afraid to fly =
We assume that the population is normally distributed
Now, the probability that the sample proportion is more than 0.11:
![P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]](https://tex.z-dn.net/?f=P%28%5Chat%7Bp%7D%3E0.11%29%3DP%28%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3E%5Cdfrac%7B0.11-0.10%7D%7B%5Csqrt%7B%5Cdfrac%7B0.10%280.90%29%7D%7B1100%7D%7D%7D%29%5C%5C%5C%5C%3DP%28z%3E%5Cdfrac%7B0.01%7D%7B0.0090453%7D%29%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%5D%5C%5C%5C%5C%3DP%28z%3E1.1055%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1.055%29%5C%5C%5C%5C%3D1-0.8543%3D0.1457%5C%20%5C%20%5C%20%5B%5Ctext%7Busing%20z-table%7D%5D)
Hence, the probability that the sample proportion is more than 0.11 = 0.1457
Answer:
range=8, mode=1 Median=3 mean=4
Step-by-step explanation:
range= subtract largest from smallest
mode= most frequent
median= middle in an ordered data set
Mean= add them all up and divide by how many you addded up
Let's compare apples to apples and oranges to oranges: convert each given proper fraction into its decimal counterpart:
3 63/80 => 3 .788 approx.
3 1/5 => 3.2
3 11/20 => 3.55
It's now an easy matter to arrange these numbers from least to greatest:
3.2, 3.55, 3.79, or
3 1/5, 3 11/20, 3 63/80
Answer:
3x - 2
Step-by-step explanation: