Answer:
1x + 4y = 4 ⇒ 6x + 24y = 24
6x - 8y = 0 ⇒ 6x - 8y = 0
32y = 24
32 32
y = 3/4
x + 4(3/4) = 4
x + 3 = 4
-3 -3
x = 1
(x, y) = (1, 3/4)
Step-by-step explanation:
up there :D
Answer:
A sample size of 128 is needed.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.93}{2} = 0.035](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.93%7D%7B2%7D%20%3D%200.035)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.81](https://tex.z-dn.net/?f=z%20%3D%201.81)
Now, find the margin of error M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population(square root of the variance) and n is the size of the sample.
How large should a sample be if the margin of error is 1 minute for a 93% confidence interval
We need a sample size of n, which is found when
. We have that
. So
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![1 = 1.81*\frac{\sqrt{39}}{\sqrt{n}}](https://tex.z-dn.net/?f=1%20%3D%201.81%2A%5Cfrac%7B%5Csqrt%7B39%7D%7D%7B%5Csqrt%7Bn%7D%7D)
![\sqrt{n} = 1.81\sqrt{39}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201.81%5Csqrt%7B39%7D)
![(\sqrt{n})^2 = (1.81\sqrt{39})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%281.81%5Csqrt%7B39%7D%29%5E2)
![n = 127.8](https://tex.z-dn.net/?f=n%20%3D%20127.8)
Rounding up
A sample size of 128 is needed.
Answer:
true false false false
Step-by-step explanation:
brainlest?
It’s full blank bro how I will see the ans bro
Answer:
The answer is 23 I just took this test and got it right
Step-by-step explanation: