Answer:
The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,

And the standard deviation of the distribution of sample mean is given by,

The information provided is:
<em>μ</em> = 144 mm
<em>σ</em> = 7 mm
<em>n</em> = 50.
Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:


*Use a <em>z</em>-table for the probability.
Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
a) 3x + 5y = 26
b) 2x + 2y = 12
First, we need to find a way to equate either the x terms or the y terms in each equation. a) 6x + 10y = 52
b) 6x + 6y = 36
Then take equation b) from the equation a) to eliminate the x component.
a) 6x + 10y = 52
- b) 6x + 6y = 36
0x + 4y = 16
y = 4 ?
Then substitute the value of y into either equation to find the value of x.
b) 2x + 2y = 12
2x + (2x4) =12
2x + 8 = 12
2x = 4
x = 2
I hope this new information i read helps you maybe understand it and as an example!
9514 1404 393
Answer:
x = 5.4
Step-by-step explanation:
The segment sum theorem tells you ...
AB +BC = AC
36 +(5x -9) = 54
5x = 27 . . . . . . . . . . . subtract 27 from both sides
x = 5.4 . . . . . . . . . . divide by 5
12 sas equals 40 over 200-30= x minus 2
X = cos(pi/6) = √3/2
y = sin(pi/6) = 1/2
So at point (√3/2, 1/2)