Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
Answer:
a) A dozen= 24
3 dozen= 24 x 3
= 72
b) 1 dozen= 12 bananas
A dozen cost= 24
So, 1 banana cost= 24/12
= 2
So, 6 bananas would be 2 x 6 = 12
c) 1 dozen= 12 bananas
A dozen cost= 24
1 banana cost= 24/12
= 2
d) I need to know how many people are there in your class so please mention that first :)
Mark me brainliest pleaseee
Use the distance formula.
Points S and W.
~3.6
Points S and T
~3.6
Points T and U
~3.1
Points U and V
~4
Points V and W
~3.3
Add all these together.
3.3 + 3.1 + 4 + 3.1 + 3.6
≈17
Answer:
i put it in the comments bc someone was answering but 150 lol
Step-by-step explanation: