The question is not complete so I will answer it with an example and a few assumptions. Follow the steps to find the answer to your question.
Important to note:
Your question is a z-scores problem
Assume that for the population of unemployed individuals the population standard deviation is 4 weeks.
Thus, we need to find the z-value.
The z-value is the sample mean decreased by the population mean, divided by the standard deviation that we assumed. So, we have:


Using any standard negative z-scores table, we can find that:
Thus, we get:

Therefore, the probability that a simple random sample of 60 unemployed individuals will provide a sample mean within 1 week of the population mean is 0.9476
Answer:
0.9476
Answer:
<em>We</em><em> </em><em>can</em><em> </em><em>say</em><em> </em><em>that</em>
<em> </em>3x + x + 8 = 32
<em>So</em><em>:</em><em> </em>
3x + x + 8 = 32
4x = 32 - 8
4x = 24
x = 24/4
<em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em>
Answer:
you should buy 2.5 okay that is more cheaper then the other ones
Step-by-step explanation:
1/2n + 9 = 27
- 9 - 9
----------------------
1/2n = 18
------ --------
1/2 1/2n
n = 36
Answer:
sec²-2scθ+tanθ+tan²θ=1-sinθ/1+sinθ
sec²-2scθ+tanθ+tan²θ
= 1+sin²θ-2sinθ/cos²θ
=-1+sinθ/1+sinθ
=1-sinθ/sinθ+1
1-sinθ/sinθ+1=-1+sinθ/1+sinθ True