1/2 (a+b)h
1/2 (a+16)14=132
8×14=112
14÷2=7
132-112=20
20÷7 =2.857142857
so then a= 2.86 (3sf)
For more questions like that turn it into equation the make a subject of formula
Well, for
her questionnaire she could use and create questions or queries that are
obviously related to her hypothesis or study.
These
could be done in a likert type of scale.
<span><span>
1.
</span>I read most often.
</span>
<span><span>a.
</span>Strongly Agree </span>
<span><span>b.
</span>Agree</span>
<span><span>c.
</span>Disagree </span>
<span><span>d.
</span>Strongly Disagree</span>
<span><span>
2.
</span>When I read my books its takes me 24 hours a day</span>
<span><span>
a.
</span>Strongly Agree </span>
<span><span>b.
</span>Agree</span>
<span><span>c.
</span>Disagree </span>
<span><span>d.
</span>Strongly Disagree</span>
<span><span>
3.
</span>When I start reading I can’t stop</span>
<span><span>
a.
</span>Strongly Agree </span>
<span><span>b.
</span>Agree</span>
<span><span>c.
</span>Disagree </span>
<span><span>d.
</span>Strongly Disagree</span>
4(m + 2) expanded is 4 x m and 4 x 2
simplified: 4m + 8
Answer:
2
Step-by-step explanation:
when it returns to the ground , h = 0
32t -16t2 = 0 and solve for t
t = 0, 2 ....0 is the start, 2 is when it returns
Answer:
c. The sampling distribution of the sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed
Step-by-step explanation:
From the given data, we have;
The category of the sample = Retired individuals
The number of participants in the sample = 20
The duration of program = six-weeks
The improvement seen by most participants = Little to no improvement
The improvement seen by few participants = Drastic improvement
Therefore, given that the participants are randomly selected and the majority of the participants make the same observation of improvement in the time to walk a mile, we have that, the majority of the outcomes show little difference in walk times after the program, therefore, the distribution of the sample data is not skewed and can be assumed to be approximately normal
