Answer:
865
Step-by-step explanation:
We have that in 95% confidence level the value of z has a value of 1.96. This can be confirmed in the attached image of the normal distribution.
Now we have the following formula:
n = [z / E] ^ 2 * (p * q)
where n is the sample size, which is what we want to calculate, "E" is the error that is 2% or 0.02. "p" is the probability they give us, 5 out of 50, is the same as 1 out of 10, that is 0.1. "q" is the complement of p, that is, 1 - 0.1 = 0.9, that is, the value of q is 0.9.
Replacing these values we are left with:
n = [1.96 / 0.02] ^ 2 * [(0.1) * (0.9)]
n = 864.36
865 by rounding to the largest number.
In the scientific value of
5.893 x 10n. The standard value is 0.00005893 what is n?
To better illustrate this
phenomenon, we can explain it further under the rules of scientific notation.
For example.
<span><span>
1. </span><span> 3 x 10^3 = 3 x
100 = 300</span></span>
<span><span>2. </span><span> 3 x 10^-3 = 3 x
0.001 = 0.003</span></span>
Solution:
0.00005893 = 5.893 x 0.00001 =
5.893 x 10^-5
n= ^-5
Answer:8:100
Step-by-step explanation:
Because the 8% is out of 100
P= M÷T , Price equals money raised divided by number of tickets sold
Looks like the given limit is
With some simple algebra, we can rewrite
then distribute the limit over the product,
The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.
For the second limit, recall the definition of the constant, <em>e</em> :
To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that
From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as
Now we apply some more properties of multiplication and limits:
So, the overall limit is indeed 0:
