The answer is 16 use pendas
Answer:
There is a 95% confidence that the true mean height of all male student at the large college is between the interval (63.5, 74.4).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean is:
![CI=\bar x\pm z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.
Or, the (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.
The 95% confidence interval for the average height of male students at a large college is, (63.5 inches, 74.4 inches).
The 95% confidence interval for the average height of male students (63.5, 74.4) implies that, there is a 0.95 probability that the true mean height of all male student at the large college is between the interval (63.5, 74.4).
Or, there is a 95% confidence that the true mean height of all male student at the large college is between the interval (63.5, 74.4).
Answer:
i wouldsay b
Step-by-step explanation:
No 8/3 is greater because if you turn them into mixed numbers 6/3 would be 2 and 8/3 would be 2 1/2
For direct variation y varies directly with x, then we use equation y = kx
Where k is the constant of proportionality
Given: Y equals 2 when X equals 3
We plug in 3 for x and 2 for y and solve for k
y = kx
2 = k(3)
Divide both sides by 3
![\frac{2}{3} =k](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%20%3Dk)
Now we plug in 2/3 for k
![y=\frac{2}{3}x](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B2%7D%7B3%7Dx)
Given x=-0.5 and find out y
![y=\frac{2}{3}(-0.5)](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B2%7D%7B3%7D%28-0.5%29)
y=-0.33
the value of y is -0.33 when x=-0.5