Answer:
A
Step-by-step explanation:
itʻll be equal to 180
the box means 90 degrees
180-90=90
90-29= 61
2*30+1=61
i hope this helps!
Answer:
0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with a mean of 2.5 feet and a standard deviation of 0.4 feet.
This means that 
Find the probability that an individual man’s step length is less than 1.9 feet.
This is the p-value of Z when X = 1.9. So



has a p-value of 0.0668
0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.
Well the P should probably be 4
Answer:
The last option
Step-by-step explanation:
To find a percentage of something you need

The expected is correct marks, the total is the total amount - then you multiply by 100 to turn the decimal into a percentage.
Answer:
a) (iii) ANOVA
b) The ANOVA test is more powerful than the t test when we want to compare group of means.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have
groups and on each group from
we have
individuals on each group we can define the following formulas of variation:



And we have this property

Solution to the problem
Part a
(i) confidence interval
False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.
(ii) t-test
Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.
(iii) ANOVA
This one is the best method when we want to compare more than 1 group of means.
(iv) Chi square
False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.
Part b
The ANOVA test is more powerful than the t test when we want to compare group of means.