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:
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:
Compute the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% as follows:
Thus, the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.
1. -4a-6=-12
add 6 to both sides
-4a-6+6=-12+6
-4a=-6
Divide by -4
2.
add
to both sides but you will need to make a common denominator on the right
Divide by
which means multiply by the <span>reciprocal</span>.
3.
Distribute
Subtract
Divide
4. 3w+7=20
Subtract
3w=13
Divide
5. -.2p+.4=-1.2
Subtract
-.2p=-1.6
Divide
p=8
6.
Add, made sure you change to common denominator first
Multiply by 4
y=5
Hope that helps
what is the question can you out it in ?
Answer:
=4.9
Step-by-step explanation:
3.4-(-1.5)
+
3.4+1.5=4.9
Answer:
-2a^4+8a^3-6a^2
Step-by-step explanation:
2a^2(a-1)(3-a)
2a^2(-a^2 +3a-3+a)
2a^2(-a^2+4a-3)
-2a^4+8a^3-6a^2