Answer:
So wavelength in decimal can be written as 0.00000155 m
Step-by-step explanation:
We have given wavelength of light used in optic fibers is ![1.55\times 10^{-6}m](https://tex.z-dn.net/?f=1.55%5Ctimes%2010%5E%7B-6%7Dm)
We have to convert this wavelength in decimal form
We know that
can be written as ![10^{-6}=\frac{1}{1000000}=0.000001](https://tex.z-dn.net/?f=10%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B1000000%7D%3D0.000001)
So ![1.55\times 10^{-6}=1.55\times 0.000001=0.00000155m](https://tex.z-dn.net/?f=1.55%5Ctimes%2010%5E%7B-6%7D%3D1.55%5Ctimes%200.000001%3D0.00000155m)
So wavelength in decimal can be written as 0.00000155 m
<u>Answer:</u>
A) 720 ways
B) 15 ways
C) 6 ways
<u>Step-by-step explanation:</u>
A) To find the number of ways Alicia can arranger her 6 paintings, we will find factorial of 6 by multiplying all of the positive integers equal to or less than that number i.e. 6 to get:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
Alicia can arrange her paintings in 720 ways.
B) We use the following formula (when order is not important) to find the number of permutations of n objects taken r at a time:
![P(n, r) = \frac{n!}{r!(n-r)!}](https://tex.z-dn.net/?f=P%28n%2C%20r%29%20%3D%20%5Cfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D)
![= \frac{6!}{4!(6-4)!} = 15](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B6%21%7D%7B4%21%286-4%29%21%7D%20%20%3D%2015)
Therefore, Alicia can choose any 4 of her paintings in 15 ways.
C) Number of ways Alicia can arrange 3 out of 6 paintings = 3! = 3*2*1 = 6 ways
Answer: = ( 7+5 ) x 1/2
Step-by-step explanation:
Answer:
The probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
![\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The sample size is, <em>n</em> = 553 > 30. Thus, the Central limit theorem is applicable in this case.
Compute the mean and standard deviation as follows:
![\mu_{\hat p}=p=0.04\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.04\times 0.96}{553}}=0.0083](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%3Dp%3D0.04%5C%5C%5C%5C%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.04%5Ctimes%200.96%7D%7B553%7D%7D%3D0.0083)
Compute the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% as follows:
![P(\hat p>0.05)=P(\frac{\hat p-\mu_{\hat p}}{\sigm_{\hat p}}>\frac{0.05-0.04}{0.0083})\\\\=P(Z>1.20)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Chat%20p%3E0.05%29%3DP%28%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigm_%7B%5Chat%20p%7D%7D%3E%5Cfrac%7B0.05-0.04%7D%7B0.0083%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.20%29%5C%5C%5C%5C%3D1-P%28Z%3C1.20%29%5C%5C%5C%5C%3D1-0.88493%5C%5C%5C%5C%3D0.11507%5C%5C%5C%5C%5Capprox%200.1151)
Thus, the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.
Answer:
2:7
Step-by-step explanation:
12:42
6:21
2:7