Answer:
For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.02 g/mi
Standard Deviation, σ = 0.01 g/mi
Sample size, n = 81
We are given that the distribution of level of nitrogen oxides is a bell shaped distribution that is a normal distribution.
Standard error due to sampling:
Formula:
We have to find the value of x such that the probability is 0.01
P(X > x)
Calculation the value from standard normal z table, we have,
For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.
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DOGE
Answer:
0.3431
Step-by-step explanation:
Here, it can work well to consider the seeds from the group of 18 that are NOT selected to be part of the group of 15 that are planted.
There are 18C3 = 816 ways to choose 3 seeds from 18 NOT to plant.
We are interested in the number of ways exactly one of the 10 parsley seeds can be chosen NOT to plant. For each of the 10C1 = 10 ways we can ignore exactly 1 parsley seed, there are also 8C2 = 28 ways to ignore two non-parsley seeds from the 8 that are non-parsley seeds.
That is, there are 10×28 = 280 ways to choose to ignore 1 parsley seed and 2 non-parsley seeds.
So, the probability of interest is 280/816 ≈ 0.3431.
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The notation nCk is used to represent the expression n!/(k!(n-k)!), the number of ways k objects can be chosen from a group of n. It can be pronounced "n choose k".
When the graph intersects the y-axis, it means it's x-coordinate is 0, or t = 0.
Hence, f(t) = 6 × 2^0 = 6.
y = 6 at y-intercept.
Answer:
Every week, the radioactivity decreases by a factor of 5
After 1 week, it goes from 5,000 to 1,000
2 weeks 1,000 / 5 =200
3 weeks 200 / 5 = 40
4 weeks 40 / 5 = 8
5 weeks 8 / 5 = 1.6
6 weeks 1.6 / 5 = 0.064
7 weeks 0.064 / 5 = 0.0128
8 weeks .0128 / 5 = 0.00256
9 weeks 0.00256 / 5 = 0.000512
10 weeks 0.000512 / 5 = 0.0001024
Step-by-step explanation:
