Answer:
The probability that the sample mean will lie within 2 values of μ is 0.9544.
Step-by-step explanation:
Here
- the sample size is given as 100
- the standard deviation is 10
The probability that the sample mean lies with 2 of the value of μ is given as

Here converting the values in z form gives

Substituting values

From z table

So the probability that the sample mean will lie within 2 values of μ is 0.9544.
Answer:
the 3rd one
Step-by-step explanation:
i just know
IF UR USERNAME IS A JOKE F U. if its not then i support you
We need a picture of the shape.
If I am reading this right, it looks like the 10, 3, 2, 1 are Adjustments and the Adjusted TB should equal the difference. Make sure you know how to add and subtract the debit and credit adjustments correctly.
TB +/- Adj = ATB
Let's draw the diagonals of the square. The square is being divided by 4 right triangles. & these right triangles are also isosceles (the diagonals of a square intercepts each other in the middle).
The area of each right triangle = (6 x 6) / 2 = 18 yard² & the 4 right triangles => 4x 18 = 72 yd²