Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.63 and standard deviation 0.80.

(a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00?

The probability that the sample average sediment density is at most 3.00 is 0.2090.

Step-by-step explanation:

Let X = he sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.63 and standard deviation, σ = 0.80.

A random sample of n = 25 specimens is selected.

The distribution of the sample mean of a normally distributed random variable is also normal.

To compute the probability of the sample mean we need to first compute the z score of the sample mean value. The formula of z-score is:

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

*Use a z-table for the probability.

Thus, the probability that the sample average sediment density is at most 3.00 is 0.2090.

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.

y=20x

: )

hope this