29 tables of relationships is the answer to your equation above … THANK ME LATER
Answer:
I agree with you,I also think its B
Step-by-step explanation:
Answer:
A is 62 b is 54 c is 62 and d is 62
And 7
Step-by-step explanation:
Since it equals to 180 you add 62 and 54 which gives u 118 then subtract it
Not sure about the second one
Answer:
Class second have higher score and have greater spread.
Step-by-step explanation:
For first box plot
For second box plot
First class has greater minimum value, it means first class has lower grades.
First quartile of both classes are same, it means equal number students in both classes have less than 62 marks.
First class has greater median.
second class has greater third quartile.
Second class has greater Maximum value. It means second class have higher score than first class.
Second class has greater range. It means the data of second class has greater spread.
Second class has greater inter quartile range. It means the data of second class has greater spread.
Therefore, the class second have higher score and have greater spread.
Answer:
a) 48.21 %
b) 45.99 %
c) 20.88 %
d) 42.07 %
e) 50 %
Note: these values represent differences between z values and the mean
Step-by-step explanation:
The test to carry out is:
Null hypothesis H₀ is μ₀ = 30
The alternative hypothesis m ≠ 30
In which we already have the value of z for each case therefore we look directly the probability in z table and carefully take into account that we had been asked for differences from the mean (0.5)
a) z = 2.1 correspond to 0.9821 but mean value is ubicated at 0.5 then we subtract 0.9821 - 0.5 and get 0.4821 or 48.21 %
b) z = -1.75 P(m) = 0.0401 That implies the probability of m being from that point p to the end of the tail, the difference between this point and the mean so 0.5 - 0.0401 = 0.4599 or 45.99 %
c) z = -.55 P(m) = 0.2912 and this value for same reason as before is 0.5 - 0.2912 = 0.2088 or 20.88 %
d) z = 1.41 P(m) = 0.9207 0.9207 -0.5 0.4207 or 42.07 %
e) z = -5.3 P(m) = 0 meaning there is not such value in z table is too small to compute and difference to mean value will be 0.5
d) z= 1.41 P(m) =
