Answer:

Step-by-step explanation:
Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.
-We calculate the probability of a mean of 3.4 as follows:
#First determine the z-value:

#We then determine the corresponding probability on the z tables:

Hence, the probability of obtaining a sample mean this large or larger is 0.0228
................
Step-by-step explanation:
it would at LEAST be like around 20-30 because a 0 can drop something FAST
Answer:
frac{21x^6y^5}{14x^2y^9}
Factor the number =\frac{7\cdot \:3x^6y^5}{14x^2y^9}
Factor the number 14=7. 2 =\frac{7\cdot \:3x^6y^5}{7\cdot \:2x^2y^9}
Cancel\:the\:common\:factor:}\:7 =\frac{3x^6y^5}{2x^2y^9}
Step-by-step explanation:
Because most of the area under any normal curve falls within a limited range of the number line