Answer:
1. 3/5
2. 70/9
3. 12 3/5
Step-by-step explanation:
hope it helps
Answer:
A sample size of at least 228 must be needed.
Step-by-step explanation:
We are given that in the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. Assume that the standard deviation is $3850.
And we have to find that what sample size do we need to have a margin of error equal to $500 with 95% confidence.
As we know that the Margin of error formula is given by;
<u>Margin of error</u> = 
where,
= significance level = 1 - 0.95 = 0.05 and
= 0.025.
= standard deviation = $3,850
n = sample size
<em>Also, at 0.025 significance level the z table gives critical value of 1.96.</em>
So, margin of error is ;
= 15.092
Squaring both sides we get,
n =
= 227.8 ≈ 228
So, we must need at least a sample size of 228 to have a margin of error equal to $500 with 95% confidence.
I think it would be C because the equation is y=7x. y=7x is also equal to x=1/7y. If y=45.50, then you would have to divide y by 7, which would equal 6.5.
A complex number takes the form
z
=
a
+
b
i
.
In this example,
a
=
0
and
b
=
3
because
z
=
0
+
3
i
.