10. A certain strain of bacteria occurs in all raw milk. Tests of uncontaminated raw milk conducted by the Health Department found that the bacteria count per milliliter is normally distributed with a mean of 3500 and standard deviation of 300. a) What is the probability of finding a bacteria count of between 3200 and 3400 in a milliliter of raw milk? b) What percentage of the uncontaminated raw milk has a bacteria count per milliliter between 3200 and 3400? c) For this normal distribution, what is the percentage of the area under the normal curve for a bacteria count per milliliter less than zero? As bacteria count cannot be less than 0, is this small enough to disregard when claiming that the bacteria count per milliliter is normally distributed? Explain your answer.

a) 0.21078

b) 21.078%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean of 3500 and standard deviation of 300.

a) What is the probability of finding a bacteria count of between 3200 and 3400 in a milliliter of raw milk?

For x = 3200

z = 3200 - 3500/300

z = -1

Probability value from Z-Table:

P(x = 3200) = 0.15866

For x = 3400

z = 3400 - 3500/300

z = -0.33333

Probability value from Z-Table:

P(x = 3400) = 0.36944

The probability of finding a bacteria count of between 3200 and 3400 in a milliliter of raw milk

P(x = 3400) - P(x = 3200)

= 0.36944 - 0.15866

= 0.21078

b) What percentage of the uncontaminated raw milk has a bacteria count per milliliter between 3200 and 3400?

We convert the answer in part a) to percentage

= 0.21078 × 100

= 21.078%

u dumb boi

step-by-step explanation:

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.