Answer:
a) 30.726m/s and b) 5.5549s
Step-by-step explanation:
a.) What was Chris Huber’s speed in meters per second(m/s)?
Given the distance and time, the formula to obtain the speed is
.
Applying this to our problem we have that
.
So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.
b) What was Whittingham’s time through the 200 m?
In a) we stated that
. This formula implies that
.
First, observer that
.
Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, 
Then, applying 1) we have that

So, Sam Whittingham’s time through the 200 m was 5.5549s.
Answer:
6.39*10^-4
Step-by-step explanation:
according to the form that is used to convert usual into scientific notation it be like 6.39*10^-4
Answer:

For the interpretation we consider a value for d small is is between 0-0.2, medium if is between 0.2-0.8 and large if is higher than 0.8.
And on this case 1.713>0.8 so we have a large effect size
This value of d=1.713 are telling to us that the two groups differ by 1.713 standard deviation and we will have a significant difference between the two means.
Step-by-step explanation:
Previous concepts
The Effect size is a "quantitative measure of the magnitude of the experimenter effect. "
The Cohen's d effect size is given by the following formula:

Solution to the problem
And for this case we can assume:
the mean for females
the mean for males
represent the deviations for both groups
And if we replace we got:

For the interpretation we consider a value for d small is is between 0-0.2, medium if is between 0.2-0.8 and large if is higher than 0.8.
And on this case 1.713>0.8 so we have a large effect size
This value of d=1.713 are telling to us that the two groups differ by 1.713 standard deviation and we will have a significant difference between the two means.
Answer: 45/8 or 5 5/8
Step-by-step explanation:
9 5 = 45
1 8 = 8
45/8=5 and 5/8
Let x be the <span>length of each of two congruent sides.
</span>The triangle will be аcute if:
x² + x² > 8²
2x² > 64
x² > 32
x > √32
x > 5.657
So, the smallest possible length of one of two congruent sides have to be 5.7 cm (<span>to the nearest tenth)</span>