Answer:
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error:
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of , so .
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is
We need a sample size of at least n, in which n is found M = 0.04.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
The answer is (x-42)×(x+42)
<span>x=theta
so we know that tanx = sinx/cosx
therefore is tanx = 4/3 and sinx = -4/5
we can set up an equation to find cosx
so 4/3 = (-4/5)/cosx
we can multiply x to have cosx(4/3)=-4/5
we can then divide 4/3 over so we have
cosx = (-4/5)/(4/3)
therefore, cos x = -3/5
</span>
there is the answer
i hope this help
Answer:
30 kilograms, I suppose. If there are 5 packages that each weight 6 kilograms, then that's 5 6 kilograms, which may equal 30 in all.
The answer is 0. if you simplify 2/4 you get 1/2. think about it. 2 parts of four is half of four, 2 is half of 4. so you get 1/2 - 1/2 which is 0
hope it helps!