27 is 144% of 18.75
mk, hope I helped
If the sample size is 300 and about 5% of the population has a particular genetic mutation then the standard deviation for the number of people with the genetic mutation in such groups of 300 is 3.77.
Given sample size be 300 and about 5% of the population has a particular genetic mutation.
We are required to find the standard deviation for the number of people with the genetic mutation in groups of 300.
Sample size=300
The standard deviation for the number of people with the genetic mutation is calculated as:
Standard deviation=
Use the known values in the above formula.
Standard deviation=
=
=3.77
Hence if the sample size is 300 and about 5% of the population has a particular genetic mutation then the standard deviation for the number of people with the genetic mutation in such groups of 300 is 3.77.
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Karen has a base salary that is $2100 and does not depends on how many copies she sells.
Additional to the base salary, she gets $50 for each copy she sells, so we can write this additional salary as 50*N, being N the number of copies she sells.
Then, we can write the total pay P as:
Then, if N=22, we can calculate the total pay as:
Answer:
The equation relating P and N is P=2100+50N.
The total pay when she sells 22 copies is $3,200.
Answer:
12x^3-9x^2
Step-by-step explanation:
Hope this helps!
Answer:
Your answer is in the screenshot! :)
Step-by-step explanation:
1. Solve for yy in y+7x=50y+7x=50. y=50-7x
y=50−7x
2. Substitute y=50-7xy=50−7x into 14x-5y=-2814x−5y=−28. 49x-250=-28
49x−250=−28
3. Solve for xx in 49x-250=-2849x−250=−28. x=\frac{222}{49}
x= 49 222
4. Substitute x=\frac{222}{49}x= 49 222 into y=50-7xy=50−7x.
y=\frac{128}{7} y= 7 128
5. Therefore, \begin{aligned}&x=\frac{222}{49}\\&y=\frac{128}{7}\end{aligned} x= 49 222 y= 7 128 Hope this helped!! :))